These NCERT Solutions for Class 12 Maths Chapter 7 समाकलन विविध प्रश्नावली Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 12 Maths Chapter 7 समाकलन विविध प्रश्नावली
प्रश्न 1.
∫\(\frac{1}{x-x^3}\)
हल:
माना I = ∫\(\frac{1}{x-x^3}\) dx
= ∫\(\frac{1}{x\left(1-x^2\right)}\) dx
1- x2 = t रखने पर तथा 1 – t = x2
– 2x dx = dt
x dx = \(-\frac{1}{2}\) dt
∴ I = – \(\frac{1}{2}\)∫\(\frac{d t}{t(1-t)}\) ……….(1)
अब \(\frac{1}{t(1-t)}\) = \(\frac{A}{t}\) + \(\frac{B}{(1-t)}\)
या
1 = A(1 – t) + Bt (आंशिक भिन्नों द्वारा)
t = 0 रखने पर,
1 = A × 1 + B × 0
∴ A = 1
t = 1 रखने पर,
1 = A(1-1) + B x 1 = B
∴ B = 1
तब (1) से,
I = – \(\frac{1}{2}\)∫(\(\frac{1}{t}\) + \(\frac{1}{2}\)∫\(\frac{1}{(1-t)}\) dt
= – \(\frac{1}{2}\)log t – \(\frac{1}{2}\)[-log(1 – t)
= – \(\frac{1}{2}\)log t + \(\frac{1}{2}\)log(1 – t)
= \(\frac{1}{2}\)log(1 – t)
∴ I = \(\frac{1}{2}\)log \(\left|\left(\frac{x^2}{1-x^2}\right)\right|\) + C
प्रश्न 2.
\(\frac{1}{\sqrt{x+a}+\sqrt{x+b}}\)
हल:
प्रश्न 3.
\(\frac{1}{x \sqrt{a x-x^2}}\)
हल:
प्रश्न 4.
\(\frac{1}{x^2\left(x^4+1\right)^{3 / 4}}\)
हल:
प्रश्न 5.
\(\frac{1}{x^{1 / 2}+x^{1 / 3}}\)
हल:
प्रश्न 6.
\(\frac{5 x}{(x+1)\left(x^2+9\right)}\)
हल:
प्रश्न 7.
\(\frac{\sin x}{\sin (x-a)}\)
हल:
माना I = ∫\(\frac{\sin x}{\sin (x-a)}\) dx
x – a = t रखने पर,
dx = dt
तथा x = t + a
∴ I = ∫\(\frac{\sin (t+a)}{\sin t}\) dt
= ∫\(\frac{\sin t \cos a+\cos t \sin a}{\sin t}\) dt
= ∫\(\frac{\sin t \cos a}{\sin t}\) dt + ∫\(\frac{\cos t}{\sin t}\) sin a dt
= cos a∫dt + sina∫cot t dt
= t cos a + sin a log|sint| + C1
= (x – a) cos a + sin a log| sin(x – a)| + C1
= sin a log|sin(x – a)| + x cos a – a cos a + C1
∴ I = sina log|sin(x – a)| + x cos a + C
(∵ C = C1 – a cos a)
प्रश्न 8.
\(\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}\)
हल:
माना I = ∫\(\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}\) dx
∴ I = ∫\(\left(\frac{\left(e^{\log x^5}-e^{\log x^4}\right)}{e^{\log x^3}-e^{\log x^2}}\right)\) dt
= ∫\(\frac{\left(x^5-x^4\right)}{\left(x^3-x^2\right)}\) dx
= ∫\(\frac{x^4(x-1)}{x^2(x-1)}\) dx
= ∫x2 dx = \(\frac{x^3}{3}\) + C
या I = \(\frac{x^3}{3}\) + C
प्रश्न 9.
\(\frac{\cos x}{\sqrt{4-\sin ^2 x}}\)
हल:
माना I = ∫\(\frac{\cos x}{\sqrt{4-\sin ^2 x}}\) dx
sin x = t
⇒ cos x dx = dt
∴ I = ∫\(\frac{d t}{\sqrt{4-t^2}}\) dt
= sin-1 \(\frac{t}{2}\) + C
∴ I = sin-1\(\left(\frac{\sin x}{2}\right)\) + C
प्रश्न 10.
\(\frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x}\)
हल:
माना I = ∫\(\frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x}\) dx
I = ∫\(\frac{\left(\sin ^4 x+\cos ^4 x\right)\left(\sin ^4 x-\cos ^4 x\right) d x}{1-2 \sin ^2 x \cos ^2 x}\) dx
= ∫\(\frac{\left(\sin ^4 x+\cos ^4 x\right)\left(\sin ^2 x+\cos ^2 x\right)}{\left(\sin ^2 x-\cos ^2 x\right)}\) dx
= \(\frac{\left(1-2 \sin ^2 x \cos ^2 x\right)\left\{-\left(\cos ^2 x-\sin ^2 x\right) \cdot 1\right\}}{\left(1-2 \sin ^2 x \cos ^2 x\right)}\) + C
∵ (sin4 x + cos4 x) = [(sin2 x) + (cos2 x)]2 – 2sin2 x cos2 x]
= – ∫cos 2x dx = –\(\frac{\sin 2 x}{2}\) + C
∴ I = –\(\frac{1}{2}\) sin 2x + C
प्रश्न 11.
\(\frac{1}{\cos (x+a) \cos (x+b)}\)
हल:
प्रश्न 12.
\(\frac{x^3}{\sqrt{1-x^8}}\)
हल:
माना I = ∫\(\frac{x^3}{\sqrt{1-\left(x^4\right)^2}}\) dx
x4 = t
⇒ 4x3dx = dt
या x3dx = \(\frac{1}{4}\) dt
∴ I = \(\frac{1}{4}\)∫\(\frac{d t}{\sqrt{1-t^2}}\)
= \(\frac{1}{4}\)sin-1 t + C
= \(\frac{1}{4}\)sin-1(x4) + C
प्रश्न 13.
\(\frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)}\)
हल:
माना I = ∫\(\frac{e^x}{\left(1+e^x\right)\left(2+e^x\right)}\) dx
ex = t
⇒ exdx = dt
∴ I = ∫\(\frac{d t}{(1+t)(2+t)}\)
अब \(\frac{1}{(1+t)(2+t)}\)
= \(\frac{A}{1+t}\) + \(\frac{B}{(2+t)}\)
1 = A(2 + t) + B(1 + t)
या 1 = 2A + At + B + Bt
या 1 = (A + B)t + 2A + B
दोनों पक्षों में t के गुणांकों तथा अचर पदों की तुलना करने पर,
A + B = 0 2A + B = 1
∴ \(\frac{1}{(1+t)(2+t)}\) = \(\frac{1}{(1+t)}\) – \(\frac{1}{(2+t)}\)
∴ I = ∫\(\frac{d t}{1+t}\) – ∫\(\frac{d t}{2+t}\)
= log|1 + t| – log|2 + t| + C
या I = log| 1 + ex| – log|2 + ex| + C
या I = log\(\left|\frac{1+e^x}{2+e^x}\right|\) + C
प्रश्न 14.
\(\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}\)
हल:
माना I = ∫\(\frac{d x}{\left(x^2+1\right)\left(x^2+4\right)}\)
पुन: x2 = y रखने पर,
∴ I = \(\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}\) = \(\frac{1}{(y+1)(y+4)}\)
= \(\frac{A}{(y+1)}\) + \(\frac{B}{(y+4)}\)
या 1 = A(-1 + 4) + B(-1 + 1)
1 = A(3) + B × 0
∴ A = \(\frac{1}{3}\)
y = – 4 रखने पर,
1 = A(-4 + 4) + B(-4 + 1)
1 = A × 0 + B(-3)
∴ B = –\(\frac{1}{3}\)
∴ \(\frac{1}{(y+1)(y+4)}\) = \(\frac{1}{3(y+1)}\) – \(\frac{1}{3(y+4)}\)
∴ I = \(\frac{1}{3}\)∫\(\frac{d x}{x^2+1}\) – \(\frac{1}{3}\)∫\(\frac{d x}{x^2+4}\)
= \(\frac{1}{3}\)tan-1x – \(\frac{1}{3}\) × \(\frac{1}{2}\) tan-1 \(\frac{x}{2}\) + C
I = \(\frac{1}{3}\)tan-1x – \(\frac{1}{6}\)tan-1 \(\frac{x}{2}\) + C
प्रश्न 15.
cos3 x elog sinx
हल:
माना I = ∫cos3 x elog sinx dx
या I = ∫cos3 x sinx dx
cosx = t रखने पर,
⇒ – sinx dx = dt
∴ I = – ∫t3 dt
या I = –\(\frac{t^4}{4}\) + C
= – \(\frac{(\cos x)^4}{4}\) + C
= –\(\frac{\cos ^4 x}{4}\) + C
∴ I = – \(\frac{1}{4}\) cos4 x + C
प्रश्न 16.
e3 logx (x4 + 1)-1
हल:
माना I = e3 logx (x4 + 1)-1 dx
या I = ∫elogx3 \(\left(\frac{1}{x^4+1}\right)\) dx
या I = ∫\(\frac{x^3}{x^4+1}\) dx
x4 + 1 = t
4x3 dx = dt
x3 dx = \(\frac{1}{4}\) dt
या I = \(\frac{1}{4}\)∫\(\frac{d t}{t}\)
= \(\frac{1}{4}\) log|t| + C
अतः I = \(\frac{1}{4}\) log|x4 + 1| + C
प्रश्न 17.
f(ax + b)[f(ax + b)]n
हल:
माना I = ∫f(ax + b)[f(ax + b)]n dx
f(ax + b) = t रखने पर,
af(ax + b) dx = dt
∴ f(ax + b) dx = \(\frac{1}{a}\) dt
I = \(\frac{1}{a}\)∫tn dt
= \(\frac{1}{a}\)\(\frac{t^{n+1}}{n+1}\) + C
अतः I = \(\frac{1}{a}\)\(\frac{[f(a x+b)]^{n+1}}{(n+1)}\) + C
प्रश्न 18.
\(\frac{1}{\sqrt{\sin ^3 x \sin (x+\alpha)}}\)
हल:
प्रश्न 19.
\(\frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}}\)
हल:
प्रश्न 20.
\(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\)
हल:
प्रश्न 21.
\(\frac{2+\sin 2 x}{1+\cos 2 x}\)
हल:
माना I = ∫\(\frac{2+\sin 2 x}{1+\cos 2 x}\) ex dx
= ∫\(\frac{2+2 \sin x \cos x}{2 \cos ^2 x}\) ex dx
= ∫\(\frac{2(1+\sin x \cos x)}{2 \cos ^2 x}\) ex dx
= ∫\(\frac{1}{\cos ^2 x}\) + \(\frac{\sin x \cos x}{\cos ^2 x}\) dx
= ∫(sec2 + tan x)ex dx
अब ex tan x = t रखने पर,
(exsec2 x + extan x) dx = dt
∴ ex(sec2x + tan x) dx = dt
∴ I = ∫dt = t + C
∴ I = extanx + C
प्रश्न 22.
\(\frac{x^2+x+1}{(x+1)^2(x+2)}\)
हल:
माना I = ∫\(\frac{x^2+x+1}{(x+1)^2(x+2)}\) dx
= \(\frac{x^2+x+1}{(x+1)^2(x+2)}\) = \(\frac{A}{x+1}\) + \(\frac{B}{(x+1)^2}\) + \(\frac{C}{(x+2)}\)
या x2 + x + 1 = A(x + 1)(x + 2) + B(x + 2) + C(x + 1)2
x = – 1
(-1)2 – 1 + 1 = A(-1 + 1)(-1 + 2) + B(-1 + 2) + C(-1 + 1)2
या 1 = B × 1 = B
∴ B = 1
x = – 2
(-2)2 – 2 + 1 = A(-2 + 1)(-2 + 2) + B(-2 + 2) + C(-2 + 1)2
या 4 – 2 + 1 = 0 + 0 + C(-1)2
या 4 – 2 + 1 = C
या 3 = C
∴ C = 3
अब दोनों पक्षों में X2 के गुणांकों की तुलना करने पर,
1 = A + C
या 1 = A + 3
या A = 1 – 3 = -2
∴ I = ∫\(\frac{x^2+x+1}{(x+1)^2(x+2)}\)
= \(\frac{-2}{x+1}\) + \(\frac{1}{(x+1)^2}\) + \(\frac{3}{(x+2)}\)
= ∫\(\frac{-2}{x+1}\) dx + ∫\(\frac{d x}{(x+1)^2}\) + 3∫\(\frac{d x}{x+2}\)
= -2 log|x + 1| + \(\frac{(x+1)^{-2+1}}{-2+1}\) + 3 log|x + 2| + C
∴ I = -2 log|x + 1| – \(\frac{1}{(x+1)}\) + 3log|x + 2| + C
प्रश्न 23.
tan-1 \(\sqrt{\frac{1-x}{1+x}}\)
हल:
प्रश्न 24.
\(\frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4}\)
हल:
प्रश्न 25.
\(\int_{\pi / 2}^\pi\) ex \(\left(\frac{1-\sin x}{1-\cos x}\right)\) dx
हल:
प्रश्न 26.
\(\int_0^{\pi / 4}\) \(\frac{\sin x \cos x}{\cos ^4 x+\sin ^4 x}\) dx
हल:
प्रश्न 27.
\(\int_0^{\pi / 2}\) \( \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x}\) dx
हल:
प्रश्न 28.
\(\int_{\pi / 6}^{\pi / 3}\) \( \frac{\sin x+\cos x}{\sqrt{\sin 2 x}}\) dx
हल:
प्रश्न 29.
\(\int_0^1 \) \(\frac{d x}{\sqrt{1+x}-\sqrt{x}}\) dx
हल:
प्रश्न 30.
\(\int_0^{\pi / 4}\)\( \frac{\sin x+\cos x}{9+16 \sin 2 x}\) dx
हल:
I = \(\int_0^{\pi / 4}\)\( \frac{\sin x+\cos x}{9+16 \sin 2 x}\) dx
sinx – cosx = t रखने पर,
(cosx + sinx) dx = dt
और (sinx – cosx)2 = t2
या sin2 x + cos2 x – 2sinx cosx = t2
या 1 – 2sinx cosx = t2
या 1 – sin2x = t2
या sin2x = 1 – t2
जब x = 0 तब sin0 – cos0 = t
या -1 = t या t = – 1
जब x = \(\frac{\pi}{4}\) तब sin\(\frac{\pi}{4}\) – cos\(\frac{\pi}{4}\) = t
या \(\frac{1}{\sqrt{2}}\) – \(\frac{1}{\sqrt{2}}\) = t
या t = 0
∴ I = \(\int_{-1}^0\) \(\frac{d t}{9+16\left(1-t^2\right)}\) = \(\int_{-1}^0\)\( \frac{d t}{25-16 t^2}\)
प्रश्न 31.
\(\int_0^{\pi / 2}\)sin2x tan-1 (sinx) dx
हल:
प्रश्न 32.
\(\int_0^\pi\) \( \frac{x \tan x}{\sec x+\tan x}\)
हल:
प्रश्न 33.
\(\int_1^4\) (|x – 1|+|x – 2|+|x – 3|)dx
हल:
प्रश्न 34.
\(\int_1^3\) \(\frac{d x}{x^2(x+1)}\) dx
हल:
प्रश्न 35.
\(\int_0^1\) xex dx = 1
हल:
माना I = \(\int_0^1\) xex dx
(x को प्रथम फलन तथा ex को द्वितीय फलन मानकर समाकलन करने पर)
अब्र \(\int_0^1\) xex dx = x∫ex dx – ∫(\(\frac{d}{d x}\) (x)∫ex dx)dx
या ∫xex dx = xex – ∫1.exdx = xex – ex
∴ I = \(\int_0^1\) xex dx = [xex – ex]
= 1(1e1 – e1) – [0e0 – e0]
= (e – e) – (-1)
= 1 R.H.S
∴ \(\int_0^1\)xex dx = 1
प्रश्न 36.
\(\int_{-1}^1\) x17cos4 dx = 0
हल:
माना \(\int_{-1}^1\) x17cos4 dx = 0
यदि f(-x) = -f(x) तब्ब फलन f विषम होगा।
∴ \(\int_{-a}^a\) f(x) dx = 0 जबकि f विषम होगा।
(-x)17 = – x17
f(-x) = (-x)17 cos4(-x)
= – x17 cos4x = -f(x)
∴ x17 cos4x विषम. है।
प्रश्न 37.
\(\int_0^{\pi / 2}\) sin3x dx = \(\frac{2}{3}\)
हल:
माना I = \(\int_0^{\pi / 2}\) sin3x dx
= \(\int_0^{\pi / 2}\) sin2x sinx dx
= \(\int_0^{\pi / 2}\) (1 – cos2x)sinx dx
cosx = t रखने पर,
– sinx dx = dt या sinx dx = – dt
जब x = 0, t = cos 0 = 1
जब x = \(\frac{\pi}{2}\) t = cos\(\frac{\pi}{2}\) = 0
∴ I = –\(\int_1^0\) (1 – t2) dt
= –\(\int_1^0\)dt + \(\int_1^0\)t2 dt
= -[t] + \(\frac{t^3}{3}\)
= -[0 – 1] + [\(\frac{0}{3}\) – \(\frac{1}{3}\)]
= 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
∴ I = \(\frac{2}{3}\)
प्रश्न 39.
\(\int_0^1\)sin-1 dx = \(\frac{\pi}{2}\) – 1
हल:
माना I = \(\int_0^1\)sin-1x dx
sin-1 x = t तब x = sin t ∴ dx = cost dt
जब x = 0 t = 0 जब x = 1 t = sin-1 = \(\frac{\pi}{2}\)
∴ I = \(\int_0^{\pi / 2}\) t.cost dt ………..(1)
अब ∫tcost dt = t∫costdt – ∫(\(\frac{d}{d t}\) t∫cost dt)
= t(sint) – ∫1.sint dt
= tsint – (-cost)
= tsint + cost
∴ I = [tsint + cost]π/2 0
= [\(\frac{\pi}{2}\)sin\(\frac{\pi}{2}\) + cos\(\frac{\pi}{2}\) – 0 × sin0 – cos0]
= \(\frac{\pi}{2}\) × 1 + 0 – 0 – 1
I = \(\frac{\pi}{2}\) × 1
प्रश्न 40.
योगफल की सीमा के रूप में \(\int_0^1\) e2-3xdx का मान ज्ञात कीजिए।
हल:
प्रश्न 41.
∫\(\frac{d x}{e^x+e^{-x}}\) बराबर है:
(A) tan-1 (ex) + C
(B) tan-1 (ex) + C
(C) log(ex – e-x) + C
(D) log(ex + ex) + C
हल:
माना I = ∫\(\frac{d x}{e^x+e^{-x}}\)
= ∫\(\frac{d x}{e^x+\frac{1}{e^x}}\) = ∫\(\frac{e^x d x}{e^{2 x}+1}\)
I = ∫\(\frac{e^x d x}{\left(e^x\right)^2+1}\), ex = t,
ex dx = dt
I = ∫\(\frac{t d t}{t^2+1}\)
= tan-1 t + C
I = tan-1(ex) + C
अतः विकल्प (A) सही है।
प्रश्न 42.
∫\(\frac{\cos 2 x}{(\sin x+\cos x)^2}\) dx बराबर है:
(A) \(\frac{-1}{\sin x+\cos x}\) + C
(B) log|sinx + cosx| + C
(C) log|sinx – cosx| + C
(D) \(\frac{1}{(\sin x+\cos x)^2}\) + C
हल:
माना I = ∫\(\frac{\cos 2 x}{(\sin x+\cos x)^2}\) dx
= ∫\(\frac{\left(\cos ^2 x-\sin ^2 x\right)}{(\sin x+\cos x)^2}\) dx
या I = ∫\( \frac{(\cos x+\sin x)(\cos x-\sin x)}{(\sin x+\cos x)^2}\) dx
I = ∫\(\frac{\cos x-\sin x}{\sin x+\cos x}\) dx
sinx + cosx = t रखने पर,
(cosx – sinx)dx = dt
∴ I = ∫\(\frac{d t}{t}\) = log|t| + C
या I = log|sinx + cosx| + C
अत: विकल्प (B) सही है।
प्रश्न 43.
यदि f(a + b – x) = f(x) तो \(\int_a^b\)x f(x) dx dx बराबर है:
(A) \(\frac{a+b}{2}\) \(\int_a^b\) f(b – x) dx
(B) \(\frac{a+b}{2}\) \(\int_a^b\) f(b + x) dx
(C) \(\left(\frac{b-a}{2}\right)\) \(\int_a^b\) f(x) dx
(D) \(\left(\frac{a+b}{2}\right)\) \(\int_a^b\) f(x) dx
हल:
माना I = ∫\(\int_a^b\) x f(x) dx
= \(\int_a^b\) (a + b – x) f(a + b – x) dx
[∴ \(\int_a^b\) f(x) dx = \(\int_a^b\) f(a + b – x) dx]
∴ I = \(\int_a^b\)(a + b – x) f(x) dx
[∵ f(a + b + x) = f(x) दिया गया है]
= \(\int_a^b\)(a + b) f(x) dx – \(\int_a^b\)x f(x) dx
= \(\int_a^b\)(a + b)f(x) dx – I
या I + I = (a + b) \(\int_a^b\) f(x) dx
या 2I = (a + b) \(\int_a^b\) f(x) dx
या I = \(\left(\frac{a+b}{2}\right)\) \(\int_a^b\) f(x) dx
अत: विकल्प (D) सही है।
प्रश्न 44.
\(\int_0^1\) tan-1 \(\left(\frac{2 x-1}{1+x-x^2}\right)\) dx बराबर है:
(A) 1
(B) 0
(C) -1
(D) \(\frac{\pi}{4}\)
हल:
