{"id":35370,"date":"2024-02-06T17:51:32","date_gmt":"2024-02-06T12:21:32","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=35370"},"modified":"2024-02-07T17:53:40","modified_gmt":"2024-02-07T12:23:40","slug":"ncert-solutions-for-class-12-maths-chapter-7-ex-7-6-in-hindi","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-7-ex-7-6-in-hindi\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 7 \u0938\u092e\u093e\u0915\u0932\u0928 Ex 7.6"},"content":{"rendered":"

These\u00a0NCERT Solutions for Class 12 Maths<\/a> Chapter 7 \u0938\u092e\u093e\u0915\u0932\u0928 Ex 7.6 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 7 \u0938\u092e\u093e\u0915\u0932\u0928 Ex 7.6<\/h2>\n

\u092a\u094d\u0930\u0936\u094d\u0928 1.
\nxsin x
\n\u0939\u0932:
\n\u092e\u093e\u0928\u093e I = \u222bx sinx dx
\n[x \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e sin x \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u0932\u0947\u0928\u0947 \u092a\u0930 ]
\n= x\u222bsin x dx – \u222b[\\(\\frac{d}{d x}\\) x \u222bsin x dx]dx
\n= x(- cos x) – \u222b[ [1.(- cos x) dx]
\n= – x cos x + \u222bcos x dx
\n= – x cos x + sin x + C<\/p>\n

\"NCERT<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 2.
\nx sin 3x
\n\u0939\u0932:
\n\u092e\u093e\u0928\u093e I = \u222bx sin 3xdx
\n[x \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e sin 3x \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u0932\u0947\u0928\u0947 \u092a\u0930]
\n= x\u222bsin 3x dx – \u222b[\\(\\frac{d}{d x}\\) x \u222bsin x dx]dx
\n= x\\(\\left(-\\frac{\\cos 3 x}{3}\\right)\\) – \u222b1\\(\\left(-\\frac{\\cos 3 x}{3}\\right)\\) dx
\n= – \\(\\frac{x}{3}\\)cos 3x + \\(\\frac{x}{3}\\)\u222bcos 3x dx
\n= – \\(\\frac{x}{3}\\)cos 3x + \\(\\frac{x}{3}\\)\\(\\frac{\\sin 3 x}{3}\\) + C
\n= – \\(\\frac{x}{3}\\)cos 3x + \\(\\frac{1}{9}\\)sin 3x + C<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 3.
\nx2<\/sup> ex<\/sup>
\n\u0939\u0932:
\n\u092e\u093e\u0928\u093e I = \u222bx2<\/sup> ex<\/sup> dx
\n(x2<\/sup> \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u092e\u093e\u0928\u0915\u0930 \u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u0930\u0928\u0947 \u092a\u0930 )
\nI = x2<\/sup>\u222bex<\/sup> dx – \u222b(\\(\\frac{d}{d x}\\)(x2<\/sup>).\u222bex<\/sup> dx)dx
\n= x2<\/sup>ex<\/sup> – \u222b2x.ex<\/sup>dx = x2<\/sup>ex<\/sup> – 2\u222bx ex<\/sup> dx
\n(\u092a\u0941\u0928\u0903 \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u092e\u093e\u0928\u0915\u0930 \u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u0930\u0928\u0947 \u092a\u0930)
\n= x2<\/sup>ex<\/sup> – 2[x\u222bex<\/sup> dx – \u222b(\\(\\frac{d}{d x}\\)(x)\u222bex<\/sup> dx)
\n= x2<\/sup>ex<\/sup> – 2[x\u222bex<\/sup> – \u222b1.ex<\/sup>dx]
\n= x2<\/sup>ex<\/sup> – 2[xex<\/sup> – ex<\/sup>] + C
\n= x2<\/sup>ex<\/sup> – 2xex<\/sup> + 2ex<\/sup> + C
\n= ex<\/sup>(x2<\/sup> – 2x + 2) + C<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 4.
\nx log x
\n\u0939\u0932:
\n\u092e\u093e\u0928\u093e I = \u222bx logx
\n(log \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e x \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u092e\u093e\u0928\u0915\u0930 \u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u0930\u0928\u0947 \u092a\u0930)
\nI = [logx \u222bxdx – \u222b{\\(\\frac{d}{d x}\\) logx\u222bx dx} dx
\n= logx \u00d7 \\(\\frac{x^2}{2}\\) – \u222b\\(\\frac{1}{x}\\) \\(\\frac{x^2}{2}\\) dx
\n= \\(\\frac{x^2}{2}\\)log x – \\(\\frac{1}{2}\\)\u222bx dx
\n= \\(\\frac{x^2}{2}\\)log x – \\(\\frac{1}{2}\\)\\(\\frac{x^2}{2}\\) + C
\n= \\(\\frac{x^2}{2}\\)log x – \\(\\frac{1}{4}\\)x2<\/sup> + C<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 5.
\nx log 2x
\n\u0939\u0932:
\n\u092e\u093e\u0928\u093e I = \u222bx log 2x dx
\n(log 2x \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u0924\u0925\u093e \u092b\u0932\u0928 \u0924\u0925\u093e x \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u092e\u093e\u0928\u0915\u0930 \u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u0930\u0928\u0947 \u092a\u0930)
\nI = [log 2x \u222bxdx – \u222b{\\(\\frac{d}{d x}\\) logx\u222bx dx} dx
\n= (log 2x) \u00d7 \\(\\frac{x^2}{2}\\) – \u222b\\(\\frac{1}{2 x}\\) \u00d7 2.\\(\\frac{x^2}{2}\\) dx
\n= \\(\\frac{x^2}{2}\\) log 2x – \\(\\frac{1}{2}\\)\u222bx dx
\n= \\(\\frac{x^2}{2}\\) log 2x – \\(\\frac{1}{2}\\)\\(\\frac{x^2}{2}\\) + C
\n= \\(\\frac{x^2}{2}\\) log 2x – \\(\\frac{1}{4}\\)x2<\/sup> + C<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 6.
\nx2<\/sup> logx
\n\u0939\u0932:
\nI = x2<\/sup> log x dx
\n(logx \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e x2 \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u092e\u093e\u0928\u0915\u0930 \u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u0930\u0928\u0947 \u092a\u0930)
\n1 = log x \u222bx2<\/sup> dx – \u222b{\\(\\frac{d}{d x}\\) logx\u222bx dx} dx
\n= (logx) \u00d7 \\(\\frac{x^3}{3}\\) – \u222b\\(\\frac{1}{x}\\) \\(\\frac{x^3}{3}\\) dx
\n= \\(\\frac{x^3}{3}\\) logx – \\(\\frac{1}{3}\\) \u222bx2<\/sup> dx
\n= \\(\\frac{x^3}{3}\\) logx – \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{x^3}{3}\\) + C
\n= \\(\\frac{x^3}{3}\\) logx – \\(\\frac{x^3}{9}\\) + C<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 7.
\nx sin-1<\/sup> x
\n\u0939\u0932:
\n\u092e\u093e\u0928\u093e
\nI = \u222bx sin-1<\/sup> x dx
\n(sin-1<\/sup> x \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e x \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u092e\u093e\u0928\u0915\u0930 \u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u0930\u0928\u0947 \u092a\u0930)
\nI = sin-1<\/sup> x \u222bxdx – \u222b{\\(\\frac{d}{d x}\\) logx\u222bx dx} dx
\n= (sin-1<\/sup> x)\\(\\frac{x^2}{2}\\) – \u222b\\(\\frac{1}{\\sqrt{1-x^2}}\\) \\(\\frac{x^2}{2}\\)
\n\"\"<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 8.
\nx tan-1<\/sup> x dx
\n\u0939\u0932:
\n\u092e\u093e\u0928\u093e
\nI = \u222bx tan-1<\/sup> x dx
\n(tan-1<\/sup> x \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e x \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u092e\u093e\u0928\u0915\u0930 \u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u0930\u0928\u0947 \u092a\u0930 )
\nI = tan-1<\/sup>x\u222bx dx – \u222b(\\(\\frac{d}{d x}\\) (tan-1<\/sup>x)\u222bxdx) dx
\n= \\(\\frac{x^2}{2}\\) tan-1<\/sup>x – \u222b\\(\\frac{1}{1+x^2}\\) \u00d7 \\(\\frac{x^2}{2}\\) dx
\n= \\(\\frac{x^2}{2}\\) tan-1<\/sup>x – \\(\\frac{1}{2}\\) \u222b\\(\\frac{x^2}{1+x^2}\\) dx
\n= \\(\\frac{x^2}{2}\\) tan-1<\/sup>x – \\(\\frac{1}{2}\\) \u222b\\(\\frac{1+x^2-1}{1+x^2}\\) dx
\n= \\(\\frac{x^2}{2}\\) tan-1<\/sup>x – \\(\\frac{1}{2}\\) \u222b\\(\\frac{\\left(1+x^2\\right)}{\\left(1+x^2\\right)}\\) dx – \\(\\frac{1}{2}\\) \u222b\\(\\frac{1}{1+x^2}\\) dx
\n= \\(\\frac{x^2}{2}\\) tan-1<\/sup>x – \\(\\frac{1}{2}\\) \u222bx dx + \\(\\frac{1}{2}\\)tan-1<\/sup>x
\n= \\(\\frac{x^2}{2}\\) tan-1<\/sup>x – \\(\\frac{1}{2}\\)x + \\(\\frac{1}{2}\\)tan-1<\/sup>x + C
\n= \\(\\frac{1}{2}\\)(x2<\/sup> + 1)tan-1<\/sup>x – \\(\\frac{1}{2}\\)x + C<\/p>\n

\"NCERT<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 9.
\nx cos-1<\/sup> x
\n\u0939\u0932:
\n\u092e\u093e\u0928\u093e I = \u222bx cos-1<\/sup> x dx
\n(cos-1<\/sup> x \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e x \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u0932\u0947\u0928\u0947 \u092a\u0930)
\nI = cos-1<\/sup>\u222bx dx – \u222b(\\(\\frac{d}{d x}\\) cos-1<\/sup> x\u222bx dx ) dx
\n= \\(\\frac{x^2}{2}\\)cos-1<\/sup>x – \u222b – \\(\\frac{1}{\\sqrt{1-x^2}}\\) \\(\\frac{x^2}{2}\\) dx
\n= \\(\\frac{x^2}{2}\\)cos-1<\/sup>x + \\(\\frac{1}{2}\\) \u222b\\(\\frac{x^2}{\\sqrt{1-x^2}}\\) dx
\n= \\(\\frac{x^2}{2}\\)cos-1<\/sup>x + \\(\\frac{1}{2}\\) I1<\/sub> ………….(1)
\nI1<\/sub> = \u222b\\(\\frac{x^2}{\\sqrt{1-x^2}}\\) dx
\n\u092e\u093e\u0928\u093e x = sin\u03b8 \u0924\u092c dx = cos\u03b8 d\u03b8
\n\"\"<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 10.
\n(sin-1<\/sup> x)2<\/sup>
\n\u0939\u0932:
\n\u092e\u093e\u0928\u093e I = \u222b(sin-1<\/sup> x)2<\/sup> dx
\n\u092e\u093e\u0928\u093e sin-1<\/sup> x = t
\n\u0924\u092c x = sint
\ndx = cost dt
\n\u2234 I = \u222b(sin-1<\/sup> x)2<\/sup> dx
\n= \u222bt2<\/sup> cos t dt
\nt2<\/sup> \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e cos t \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u092e\u093e\u0928\u0915\u0930 \u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u0930\u0928\u0947 \u092a\u0930)
\nI = t2<\/sup>\u222bcost dt – \u222b(\\(\\frac{d}{d t}\\)(t2<\/sup>) \u222bcos t dt) dt
\nt2<\/sup>sint – \u222b2tsint dt
\n= t2<\/sup>sint – 2\u222btsint dt
\nI = t2<\/sup>sint – 2I1<\/sub> ……………….(1)
\n\u2234 I1<\/sub> = \u222bt sint dt – \u222b(\\(\\frac{d}{d t}\\) (t) \u222bsint dt) dt
\n( t \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e sin t \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u0932\u0947\u0928\u0947 \u092a\u0930)
\n= t(-cost) – \u222b1.(-cost)dt
\n= -tcost + \u222bcos t dt
\n= -t cost + sin t + C1<\/sub>
\nI1<\/sub> \u0915\u093e \u092e\u093e\u0928 \u0938\u092e\u0940\u0915\u0930\u0923 (1) \u092e\u0947\u0902 \u0930\u0916\u0928\u0947 \u092a\u0930,
\nI = t2<\/sup> sint – 2(-t cost + sin t + C1<\/sub>)
\n= t2<\/sup> sin t + 2t cost – 2 sint – 2C1<\/sub>
\n= t2<\/sup> sin t + 2t cost – 2 sint + C (\u2235 C = -2C1<\/sub>)
\n= x(sin-1<\/sup> x)2<\/sup> + 2 sin-1<\/sup>\\(x \\sqrt{1-x^2}\\) – 2x + C<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 11.
\n\\(\\frac{x \\cos ^{-1} x}{\\sqrt{1-x^2}}\\)
\n\u0939\u0932:
\n\u092e\u093e\u0928\u093e I = \u222b\\(\\frac{x \\cos ^{-1} x}{\\sqrt{1-x^2}}\\) dx
\ncos-1<\/sup> x = t \u0930\u0916\u0928\u0947 \u092a\u0930
\n\u21d2 x = cos t
\n(t \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e cost \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u092e\u093e\u0928\u0915\u0930 \u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u0930\u0928\u0947 \u092a\u0930)
\nI = -[t \u222bcos t dt – \u222b{\\(\\frac{d}{d t}\\) t \u222bcost dt}dt
\n=- [t sint – \u222b1.sin t dt] = -t sin t + \u222bsin t dt
\n= -tsin t + (-cost) + C
\n= -tsint – cost + C
\n= -t\\(\\sqrt{1-\\cos ^2 t}\\) – cost + C
\n= – (cos-1<\/sup>x)\\(\\sqrt{1-x^2}\\) – x + C<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 12.
\nx sec2<\/sup> x
\n\u0939\u0932:
\n\u092e\u093e\u0928\u093e
\nI = \u222bx sec2<\/sup> x dx
\n(x \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u0924\u0925\u093e sec2<\/sup> x \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u092e\u093e\u0928\u0915\u0930 \u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u0930\u0928\u0947 \u092a\u0930)
\nI = x\u222bsec2<\/sup> x dx – \u222b(\\(\\frac{d}{d x}\\) x \u222bsec2<\/sup> x dx)dx
\n= x tan x – \u222b1.tan x dx
\n= x tan x – \u222btanx dx
\n= xtan x – (-log|cos x|) + C
\n= xtanx + log|cosx| + C<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 13.
\ntan-1<\/sup> x
\n\u0939\u0932:
\n\u092e\u093e\u0928\u093e
\nI = \u222btan-1<\/sup> x dx
\n= \u222b(tan-1<\/sup> x ) 1. dx
\n(tan-1<\/sup> x \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e 1 \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u092e\u093e\u0928\u0915\u0930 \u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u0930\u0928\u0947 \u092a\u0930)
\nI = tan-1<\/sup> x\u222b1 dx – \u222b(\\(\\frac{d}{d x}\\) tan-1<\/sup> x \u222b1 dx)dx
\n= x tan-1<\/sup> x – \u222b\\(\\frac{1}{1+x^2}\\) x dx
\n= x tan-1<\/sup> x – \u222b\\( \\frac{x}{1+x^2}\\) dx
\nI = x tan-1<\/sup> x – I1<\/sub>(\u092e\u093e\u0928\u093e) ………….(1)
\n\u2234 I1<\/sub> = \u222b\\( \\frac{x}{1+x^2}\\) dx
\n\u092e\u093e\u0928\u093e 1 + x2<\/sup> = t
\n\u0924\u092c 2x dx = dt
\n\u2234 xdx = \\(\\frac{1}{2}\\) dt
\nI1<\/sub> = \\(\\frac{1}{2}\\)\u222b\\(\\frac{d t}{t}\\) = \\(\\frac{1}{2}\\)log|t| + C1<\/sub>
\n= \\(\\frac{1}{2}\\)log|1 + x2<\/sup>| + C1<\/sub>
\nI1<\/sub> \u0915\u093e \u092e\u093e\u0928 (1) \u092e\u0947\u0902 \u0930\u0916\u0928\u0947 \u092a\u0930,
\nI = xtan-1<\/sup> x – {\\(\\frac{1}{2}\\)log|1 + x2<\/sup> | – C1<\/sub>
\n= xtan-1<\/sup> x – \\(\\frac{1}{2}\\)log|1 + x2<\/sup> | + C
\n= xtan-1<\/sup> x – \\(\\frac{1}{2}\\)log(1 + x2<\/sup>) + C
\n(\u2235 x2<\/sup> > 0 1 > 0, \u2234 1 + x2<\/sup> > 0)<\/p>\n

\"NCERT<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 14.
\nx (logx)2<\/sup>
\n\u0939\u0932:
\n\u092e\u093e\u0928\u093e I = \u222bx (logx)2<\/sup> dx
\n((logx)2<\/sup> \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u0924\u0925\u093e x \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u092e\u093e\u0928\u0915\u0930 \u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u0930\u0928\u0947 \u092a\u0930)
\nI = (logx)2<\/sup>\u222bx dx – \u222b(\\(\\frac{d}{d x}\\)(logx)2<\/sup>\u222bx dx)dx
\n= \\(\\frac{x^2}{2}\\)(logx)2<\/sup> – \u222b\\(\\frac{2 \\log x}{x}\\) \u00d7 \\(\\frac{x^2}{2}\\) dx
\n= \\(\\frac{x^2}{2}\\)(logx)2<\/sup> – \u222bxlogx dx
\nI = \\(\\frac{x^2}{2}\\)(logx)2<\/sup> – I1<\/sub>(\u092e\u093e\u0928\u093e) …………….(1)
\n\"\"<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 15.
\n(x2<\/sup> + 1)logx
\n\u0939\u0932:
\n\u092e\u093e\u0928\u093e I = \u222b(x2<\/sup> + 1)logx dx
\n(logx \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e (x2<\/sup> + 1) \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u092e\u093e\u0928\u0915\u0930 \u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u0930\u0928\u0947 \u092a\u0930)
\nI = logx\u222b(x2<\/sup> + 1)dx – \u222b(\\(\\frac{d}{d x}\\) (logx) \u222b(x2<\/sup> + 1)dx
\n= logx(\\(\\frac{x^3}{3}\\) + x) – \u222b\\(\\frac{1}{x}\\) (\\(\\frac{x^3}{3}\\) + x)dx
\n= (\\(\\frac{x^3}{3}\\) + x)logx – \u222b\\(\\frac{1}{3x}\\)(\\(\\frac{x^3}{3}\\) + x)dx
\n= \\(\\left(\\frac{3 x+x^3}{3}\\right)\\) logx – \\(\\frac{1}{3}\\)\u222bx2<\/sup>dx – \u222bdx
\n= \\(\\left(\\frac{3 x+x^3}{3}\\right)\\) logx – \\(\\frac{1}{3}\\)\\(\\frac{x^3}{3}\\) – x + C
\n= \\(\\frac{1}{3}\\)(3x + x3<\/sup>) logx – \\(\\frac{x^3}{9}\\) – x + C<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 16.
\nex<\/sup> (sinx + cosx)
\n\u0939\u0932:
\n\u092e\u093e\u0928\u093e I = \u222bex<\/sup> (sinx + cosx) dx
\n= \u222bex<\/sup>sinx + \u222bex<\/sup>cosx dx
\n(\u092a\u094d\u0930\u0925\u092e \u0938\u092e\u093e\u0915\u0932\u0928 \u092e\u0947\u0902 sin x \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e ex<\/sup> \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u092e\u093e\u0928\u0915\u0930 \u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u0930\u0928\u0947 \u092a\u0930)
\nI = sinx\u222bex<\/sup> dx – \u222b\\(\\frac{d}{d x}\\) sinx\u222bex<\/sup> dx + \u222bex<\/sup>cosx dx
\n= ex<\/sup>sinx – \u222b(cosx)ex<\/sup>dx + \u222bex<\/sup>cosxdx + C
\n= ex<\/sup>sinx – \u222bex<\/sup>cosxdx + \u222bex<\/sup>cosxdx + C
\n= ex<\/sup>sinx + C<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 17.
\n\\(\\frac{x e^x}{(1+x)^2}\\)
\n\u0939\u0932:
\n\"\"<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 18.
\nex<\/sup> \\(\\left(\\frac{1+\\sin x}{1+\\cos x}\\right)\\)
\n\u0939\u0932:
\nI = \u222bex<\/sup> \\(\\left(\\frac{1+\\sin x}{1+\\cos x}\\right)\\) dx
\n= \u222bex<\/sup> \\(\\frac{(1+2 \\sin x \/ 2 \\cos x \/ 2)}{2 \\cos ^2 x \/ 2}\\) dx
\n[\u2235 sinx = 2 sin x \/ 2cosx \/ 2 \u0924\u0925\u093e cos x = 2 cos2<\/sup> x \/ 2 – 1]
\n= \u222bex<\/sup>(\\(\\frac{1}{2}\\) sec2<\/sup> \\(\\frac{x}{2}\\) + tan\\(\\frac{x}{2}\\))dx
\n= \u222bex<\/sup>tan\\(\\frac{x}{2}\\)dx + \\(\\frac{1}{2}\\)\u222bex<\/sup>sec2<\/sup> \\(\\frac{x}{2}\\)dx
\n(\u092a\u094d\u0930\u0925\u092e \u0938\u092e\u093e\u0915\u0932\u0928 \u092e\u0947\u0902 tan\\(\\frac{x}{2}\\) \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e ex<\/sup> \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u092e\u093e\u0928\u0915\u0930 \u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u0930\u0928\u0947 \u092a\u0930)
\nI = [tan\\(\\frac{x}{2}\\)\u222bex<\/sup>dx – \u222b(\\(\\frac{d}{d x}\\)tan\\(\\frac{x}{2}\\)dx\u222bex<\/sup>dx)dx + \\(\\frac{1}{2}\\)\u222bex<\/sup>sec2<\/sup> \\(\\frac{x}{2}\\) dx + C
\n= [tan\\(\\frac{x}{2}\\) ex<\/sup> – \u222b\\(\\frac{1}{2}\\) sec2<\/sup>\\(\\frac{x}{2}\\) dx + C
\n= ex<\/sup> tan\\(\\frac{x}{2}\\) – \\(\\frac{1}{2}\\)\u222bex<\/sup>sec2<\/sup> \\(\\frac{x}{2}\\) dx + C
\n= ex<\/sup> tan\\(\\frac{x}{2}\\) + C<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 19.
\nex<\/sup>(\\(\\frac{1}{x}\\) – \\(\\frac{1}{x^2}\\))
\n\u0939\u0932:
\nI = \u222bex<\/sup>(\\(\\frac{1}{x}\\) – \\(\\frac{1}{x^2}\\)) dx
\n\u092f\u093e I = \u222bex<\/sup>\\(\\frac{1}{x}\\) dx – \u222bex<\/sup>\\(\\frac{1}{x^2}\\) dx
\n(\u092a\u094d\u0930\u0925\u092e \u0938\u092e\u093e\u0915\u0932\u0928 \u092e\u0947\u0902 \\(\\frac{1}{x}\\) \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e ex<\/sup> \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u092e\u093e\u0928\u0915\u0930 \u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u0930\u0928\u0947 \u092a\u0930)
\nI = \\(\\frac{1}{x}\\)\u222bex<\/sup>dx – \u222b(\\(\\frac{d}{d x}\\)\u222bex<\/sup>dx)dx – \u222bex<\/sup> \\(\\frac{1}{x^2}\\)dx
\n= \\(\\frac{1}{x}\\)\u222bex<\/sup> – \u222b-\\(\\frac{1}{x^2}\\)ex<\/sup>dx – \u222bex<\/sup> \\(\\frac{1}{x^2}\\)dx
\n= \\(\\frac{1}{x}\\)\u222bex<\/sup> + \u222bex<\/sup>\\(\\frac{1}{x^2}\\)dx – \u222bex<\/sup>\\(\\frac{1}{x^2}\\)dx + C
\n= \\(\\frac{1}{x}\\)\u222bex<\/sup> + C<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 20.
\n\\(\\frac{(x-3) e^x}{(x-1)^3}\\)
\n\u0939\u0932:
\nI = \u222b\\(\\frac{(x-3) e^x}{(x-1)^3}\\) dx
\n= \u222b\\(\\frac{e^x(x-1-2)}{(x-1)^3}\\) dx
\n= \u222bex<\/sup>(\\(\\frac{(x-1)}{(x-1)^3}\\) – \\(\\frac{2}{(x-1)^3}\\))dx
\n= \u222bex<\/sup>(\\(\\frac{1}{(x-1)^2}\\) – \\(\\frac{2}{(x-1)^3}\\))dx
\n\u092f\u0939\u093e\u0901 f(x) = \\(\\frac{1}{(x-1)^2}\\) \u0924\u0925\u093e f(x) = –\\(\\frac{2}{(x-1)^3}\\)
\n\u0924\u092c \u0938\u0942\u0924\u094d\u0930 \u222bex<\/sup>[f(x) + f(x)]dx = ex<\/sup> f(x) \u0938\u0947
\nI = \\(\\frac{e^x}{(x-1)^2}\\) + C<\/p>\n

\"NCERT<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 21.
\ne2x<\/sup> sinx
\n\u0939\u0932:
\nI = \u222be2x<\/sup> sinx dx
\n(e2x<\/sup> \u092a\u094d\u0930\u0925\u092e \u0938\u092e\u093e\u0915\u0932\u0928 \u092e\u0947\u0902 sinx \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u092e\u093e\u0928\u0915\u0930 \u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u0930\u0928\u0947 \u092a\u0930)
\nI = \u222be2x<\/sup>\u222bsinx dx – \u222b(\\(\\frac{d}{d x}\\)e2x<\/sup> \u222bcosx dx)dx
\n\u092f\u093e I1<\/sub> = e2x<\/sup>sinx – \u222b2e2x<\/sup>sinx dx + C1<\/sub>
\nI1<\/sub> \u0915\u093e \u092e\u093e\u0928 \u0938\u092e\u0940\u0915\u0930\u0923 (1) \u092e\u0947\u0902 \u0930\u0916\u0928\u0947 \u092a\u0930,
\nI = -e2x<\/sup>cosx + 2[e2x<\/sup> sinx – 2\u222be2x<\/sup>sinx – 2\u222be2x<\/sup>sinx dx + C1<\/sub>
\n\u092f\u093e
\nI = -e2x<\/sup>cosx + 2e2x<\/sup> sinx – 4\u222be2x<\/sup>sinxdx + 2C1<\/sub>
\n\u092f\u093e
\nI = -e2x<\/sup>cosx + 2e2x<\/sup> sinx – 4I + 2C1<\/sub>
\n\u092f\u093e
\nI + 4I = -e2x<\/sup>cosx + 2e2x<\/sup> sinx + 2C1<\/sub>
\n\u092f\u093e
\n5I = -e2x<\/sup>cosx + 2e2x<\/sup> sinx + 2C1<\/sub>
\n\u092f\u093e
\nI = –\\(\\frac{1}{5}\\) e2x<\/sup>cosx + \\(\\frac{2}{5}\\)2e2x<\/sup> sinx + \\(\\frac{2}{5}\\)C1<\/sub>
\n\u092f\u093e
\nI = –\\(\\frac{1}{5}\\) e2x<\/sup>cosx + \\(\\frac{2}{5}\\)e2x<\/sup> sinx + C
\n(\u2235 C = \\(\\frac{2}{5}\\)C1<\/sub>)
\nI = \\(\\frac{e^{2 x}}{5}\\) (2sinx – cosx) + C<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 22.
\nsin-1<\/sup> \\(\\left(\\frac{2 x}{1+x^2}\\right)\\)
\n\u0939\u0932:
\nI = \u222bsin-1<\/sup> \\(\\left(\\frac{2 x}{1+x^2}\\right)\\) dx
\n\u092e\u093e\u0928
\nx = tan\u03b8 \u0924\u092c \u03b8 = tan-1<\/sup> x
\n\u0924\u0925\u093e dx = sec2<\/sup> \u03b8 d\u03b8
\nI = \u222bsin-1<\/sup>\\(\\left(\\frac{2 \\tan \\theta}{1+\\tan ^2 \\theta}\\right)\\) sec2<\/sup> \u03b8 d\u03b8
\n= \u222bsin-1<\/sup> (sin2\u03b8) sec2<\/sup> \u03b8 d\u03b8
\n[\u2235 sin2\u03b8 = \\(\\frac{2 \\tan \\theta}{1+\\tan ^2 \\theta}\\)]
\n= \u222b2\u03b8 sec2<\/sup> \u03b8 d\u03b8 = 2\u222b\u03b8sec2<\/sup> \u03b8 d\u03b8
\n(\u03b8 \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e sec2<\/sup> \u03b8 \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u092e\u093e\u0928\u0915\u0930 \u0938\u092e\u093e\u0915\u0932\u0928 \u0915\u0930\u0928\u0947 \u092a\u0930)
\nI = 2[\u03b8\u222bsec2<\/sup> \u03b8 d\u03b8 – \u222b(\\(\\frac{d}{d \\theta}\\)\u222bsec2<\/sup> \u03b8 d\u03b8)d\u03b8
\n= 2\u03b8tan\u03b8 – 2\u222b1.tan\u03b8d\u03b8
\n= 2\u03b8tan\u03b8 – 2(-log|cos\u03b8|) v + C
\n\"\"
\n= 2xtan-1<\/sup> x + 2log\\(\\left|\\frac{1}{\\sqrt{1+x^2}}\\right|\\) + C
\n= 2xtan-1<\/sup> x + 2log|(1 + x2<\/sup> )-1\/2<\/sup>| + C
\n= 2xtan-1<\/sup> x + 2(-1\/2)log|(1 + x2<\/sup>)| + C
\n= 2xtan-1<\/sup> x – log(1 + x2<\/sup>) + C<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 23.
\n\u222bx2<\/sup>ex3<\/sup> dx
\n(A) \\(\\frac{1}{3}\\) ex3<\/sup> + C
\n(B) \\(\\frac{1}{3}\\)ex2<\/sup> + C
\n(C) \\(\\frac{1}{2}\\)ex3<\/sup> + C
\n(D) \\(\\frac{1}{2}\\)ex2<\/sup> + C
\n\u0939\u0932:
\nI = \u222bx2<\/sup>ex3<\/sup> dx
\n\u092e\u093e\u0928\u093e x3<\/sup> = t
\n\u0924\u092c 3x2<\/sup> = dt
\n\u092f\u093e x2<\/sup>dx = dt
\n\u2234 I = \\(\\frac{1}{3}\\)\u222bet<\/sup> + C
\n= \\(\\frac{1}{3}\\)\u222bex3<\/sup> + C
\n\u0905\u0924\u0903 \u0935\u093f\u0915\u0932\u094d\u092a (A) \u0938\u0939\u0940 \u0939\u0948\u0964<\/p>\n

\"NCERT<\/p>\n

\u092a\u094d\u0930\u0936\u094d\u0928 24.
\n\u222bex<\/sup> x(1 + tanx)dx
\n(A) ex<\/sup>cosx + C
\n(B) ex<\/sup>secx + C
\n(C) ex<\/sup>sinx + C
\n(D) ex<\/sup>tanx + C
\n\u0939\u0932:
\nI = \u222bex<\/sup>secx(1 + tanx) dx
\n= \u222bex<\/sup>(secx + secx tanx)dx
\n\u092f\u0939\u093e\u0901 f(x) = secx \u0924\u0925\u093e f(x) = secxtanx
\n\u0924\u092c \u0938\u0942\u0924\u094d\u0930 \u222bex<\/sup>[f(x) + f(x)]dx = ex<\/sup>f(x) + C
\nI = ex<\/sup>secx + C
\n\u0905\u0924\u0903 \u0935\u093f\u0915\u0932\u094d\u092a (B) \u0938\u0939\u0940 \u0939\u0948\u0964<\/p>\n","protected":false},"excerpt":{"rendered":"

These\u00a0NCERT Solutions for Class 12 Maths Chapter 7 \u0938\u092e\u093e\u0915\u0932\u0928 Ex 7.6 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 12 Maths Chapter 7 \u0938\u092e\u093e\u0915\u0932\u0928 Ex 7.6 \u092a\u094d\u0930\u0936\u094d\u0928 1. xsin x \u0939\u0932: \u092e\u093e\u0928\u093e I = \u222bx sinx dx [x \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e sin x \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u0932\u0947\u0928\u0947 …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 7 \u0938\u092e\u093e\u0915\u0932\u0928 Ex 7.6<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 7 \u0938\u092e\u093e\u0915\u0932\u0928 Ex 7.6 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-7-ex-7-6-in-hindi\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 12 Maths Chapter 7 \u0938\u092e\u093e\u0915\u0932\u0928 Ex 7.6 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These\u00a0NCERT Solutions for Class 12 Maths Chapter 7 \u0938\u092e\u093e\u0915\u0932\u0928 Ex 7.6 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 12 Maths Chapter 7 \u0938\u092e\u093e\u0915\u0932\u0928 Ex 7.6 \u092a\u094d\u0930\u0936\u094d\u0928 1. xsin x \u0939\u0932: \u092e\u093e\u0928\u093e I = \u222bx sinx dx [x \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e sin x \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u0932\u0947\u0928\u0947 … NCERT Solutions for Class 12 Maths Chapter 7 \u0938\u092e\u093e\u0915\u0932\u0928 Ex 7.6 Read More »\" \/>\n<meta property=\"og:url\" content=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-7-ex-7-6-in-hindi\/\" \/>\n<meta property=\"og:site_name\" content=\"NCERT Solutions\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/NCERTSolutionsGuru\/\" \/>\n<meta property=\"article:published_time\" content=\"2024-02-06T12:21:32+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2024-02-07T12:23:40+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&ssl=1\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:creator\" content=\"@ncertsolguru\" \/>\n<meta name=\"twitter:site\" content=\"@ncertsolguru\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Prasanna\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"12 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Organization\",\"@id\":\"https:\/\/ncertsolutions.guru\/#organization\",\"name\":\"NCERT Solutions\",\"url\":\"https:\/\/ncertsolutions.guru\/\",\"sameAs\":[\"https:\/\/www.facebook.com\/NCERTSolutionsGuru\/\",\"https:\/\/www.instagram.com\/ncertsolutionsguru\/\",\"https:\/\/www.linkedin.com\/in\/ncertsolutionsguru\/\",\"https:\/\/in.pinterest.com\/ncertsolutionsguru\/\",\"https:\/\/en.wikipedia.org\/wiki\/National_Council_of_Educational_Research_and_Training\",\"https:\/\/twitter.com\/ncertsolguru\"],\"logo\":{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/ncertsolutions.guru\/#logo\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1\",\"contentUrl\":\"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1\",\"width\":460,\"height\":45,\"caption\":\"NCERT Solutions\"},\"image\":{\"@id\":\"https:\/\/ncertsolutions.guru\/#logo\"}},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/ncertsolutions.guru\/#website\",\"url\":\"https:\/\/ncertsolutions.guru\/\",\"name\":\"NCERT Solutions\",\"description\":\"NCERT Solutions for Class 1 to 12\",\"publisher\":{\"@id\":\"https:\/\/ncertsolutions.guru\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/ncertsolutions.guru\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-7-ex-7-6-in-hindi\/#primaryimage\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&ssl=1\",\"contentUrl\":\"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&ssl=1\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-7-ex-7-6-in-hindi\/#webpage\",\"url\":\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-7-ex-7-6-in-hindi\/\",\"name\":\"NCERT Solutions for Class 12 Maths Chapter 7 \\u0938\\u092e\\u093e\\u0915\\u0932\\u0928 Ex 7.6 - 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NCERT Solutions for Class 12 Maths Chapter 7 \u0938\u092e\u093e\u0915\u0932\u0928 Ex 7.6 \u092a\u094d\u0930\u0936\u094d\u0928 1. xsin x \u0939\u0932: \u092e\u093e\u0928\u093e I = \u222bx sinx dx [x \u0915\u094b \u092a\u094d\u0930\u0925\u092e \u092b\u0932\u0928 \u0924\u0925\u093e sin x \u0915\u094b \u0926\u094d\u0935\u093f\u0924\u0940\u092f \u092b\u0932\u0928 \u0932\u0947\u0928\u0947 … NCERT Solutions for Class 12 Maths Chapter 7 \u0938\u092e\u093e\u0915\u0932\u0928 Ex 7.6 Read More »","og_url":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-7-ex-7-6-in-hindi\/","og_site_name":"NCERT Solutions","article_publisher":"https:\/\/www.facebook.com\/NCERTSolutionsGuru\/","article_published_time":"2024-02-06T12:21:32+00:00","article_modified_time":"2024-02-07T12:23:40+00:00","og_image":[{"url":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&ssl=1"}],"twitter_card":"summary_large_image","twitter_creator":"@ncertsolguru","twitter_site":"@ncertsolguru","twitter_misc":{"Written by":"Prasanna","Est. reading time":"12 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Organization","@id":"https:\/\/ncertsolutions.guru\/#organization","name":"NCERT Solutions","url":"https:\/\/ncertsolutions.guru\/","sameAs":["https:\/\/www.facebook.com\/NCERTSolutionsGuru\/","https:\/\/www.instagram.com\/ncertsolutionsguru\/","https:\/\/www.linkedin.com\/in\/ncertsolutionsguru\/","https:\/\/in.pinterest.com\/ncertsolutionsguru\/","https:\/\/en.wikipedia.org\/wiki\/National_Council_of_Educational_Research_and_Training","https:\/\/twitter.com\/ncertsolguru"],"logo":{"@type":"ImageObject","@id":"https:\/\/ncertsolutions.guru\/#logo","inLanguage":"en-US","url":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1","contentUrl":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1","width":460,"height":45,"caption":"NCERT Solutions"},"image":{"@id":"https:\/\/ncertsolutions.guru\/#logo"}},{"@type":"WebSite","@id":"https:\/\/ncertsolutions.guru\/#website","url":"https:\/\/ncertsolutions.guru\/","name":"NCERT Solutions","description":"NCERT Solutions for Class 1 to 12","publisher":{"@id":"https:\/\/ncertsolutions.guru\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/ncertsolutions.guru\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"ImageObject","@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-7-ex-7-6-in-hindi\/#primaryimage","inLanguage":"en-US","url":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&ssl=1","contentUrl":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?resize=170%2C17&ssl=1"},{"@type":"WebPage","@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-7-ex-7-6-in-hindi\/#webpage","url":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-7-ex-7-6-in-hindi\/","name":"NCERT Solutions for Class 12 Maths Chapter 7 \u0938\u092e\u093e\u0915\u0932\u0928 Ex 7.6 - 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