NCERT Solutions for Class 12 Maths Chapter 7 समाकलन Ex 7.10

These NCERT Solutions for Class 12 Maths Chapter 7 समाकलन Ex 7.10 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 12 Maths Chapter 7 समाकलन Ex 7.10

प्रश्न 1.
\(\int_0^1\) \(\frac{x}{x^2+1}\) dx
हल:
माना I = \(\int_0^1\) \(\frac{x}{x^2+1}\) dx
x² + 1 = t रखने पर,
⇒ 2x dx = dt ⇒ x dx = \(\frac{1}{2}\) dt
जब x = 0, t = 1, x = 1 तब 1 = 1² + 1 = 2
∴ I = \(\int_0^1\) \(\frac{x}{x^2+1}\) dx
= \(\frac{1}{2}\) \(\int_2^1\) \(\frac{d t}{t}\)
= \(\frac{1}{2}\)[log t]² ₁
= \(\frac{1}{2}\)[log2 – log1]
∴ I = \(\frac{1}{2}\)log 2

प्रश्न 2.
\(\int_0^{\pi / 2}\) √sinΦcos⁵Φ dΦ
हल:

प्रश्न 3.
\(\int_0^1\) sin^-1(\(\frac{2 x}{1+x^2}\))dx
हल:
माना
I = \(\int_0^1\) sin^-1(\(\frac{2 x}{1+x^2}\))dx
x = tanθ रखने पर, dx = sec²θ dθ
जब x = 1, θ = \(\frac{\pi}{4}\) जब x = 0, θ = 0
[∴ tan θ = 1 = tan \(\frac{\pi}{4}\) ⇒ θ = \(\frac{\pi}{4}\)]
∴ I = \(\int_0^{\pi / 4}\) sin^-1(\(\frac{2 \tan \theta}{1+\tan ^2 \theta}\) ) sec²θ dθ
I = \(\int_0^{\pi / 4}\) sin^-1 (sin2θ) sec²θ dθ
[∵ sin2A = \(\frac{2 \tan A}{1+\tan ^2 A}\)]
= \(\int_0^{\pi / 4}\) 2θ sec²θ dθ
I = 2\(\int_0^{\pi / 4}\) θ sec²θ dθ ………….(1)
∫θ sec²θ dθ में θ को प्रथम फलन तथा sec²θ को द्वितीय फलन लेकर समाकलन करने पर,
∫θ sec²θ dθ
= θ ∫sec²θ dθ
= θ ∫sec²θ – ∫[\(\frac{d}{d \theta}\) (θ) ∫sec²θ dθ] dθ
= θtanθ – ∫1. tanθ dθ
= θtanθ – (-log|cosθ|)
= θtanθ + log|cosθ|
∴ ∫θ sec²θ dθ = θ tanθ + log|cosθ|
∫θ sec²θ dθ का मान (1) में रखने पर,

प्रश्न 4.
\(\int_0^1\) \(x \sqrt{x+2}\) dx
हल:
माना I = \(\int_0^1\) \(x \sqrt{x+2}\) dx
x + 2 = t² रखने पर, dx = 2t dt
जब x = 0, तब t = √2, x = 2 तब t² = 2 + 2 = 4 t = 2

प्रश्न 5.
\(\int_0^{\pi / 2}\) \(\frac{\sin x}{1+\cos ^2 x}\) dx
हल:
माना I = \(\int_0^{\pi / 2}\) \(\frac{\sin x}{1+\cos ^2 x}\) dx
cosx = t रखने पर,
– sinx dx = dt, sinx dx = -dt
x = 0, t = cosθ = 1, x = \(\frac{\pi}{2}\), t = cos\(\frac{\pi}{2}\) = 0
∴ I = \(\int_1^0\) \(\frac{-d t}{1+t^2}\) = –\(\int_1^0\) \(\frac{d t}{1+t^2}\) = -tan^-1 t
= -[tan^-1 t]⁰ ₁ = – [tan^-10 – tan^-11]
= -[0 – \(\frac{\pi}{4}\)] = \(\frac{\pi}{4}\)
∴ I = \(\frac{\pi}{4}\)

प्रश्न 6.
\(\int_0^2\) \(\frac{d x}{x+4-x^2}\)
हल:

प्रश्न 7.
\(\int_{-1}^1\) \(\frac{d x}{x^2+2 x+5}\)
हल:

प्रश्न 8.
\(\int_1^2\) (\(\frac{1}{x}\) – \(\frac{1}{2 x^2}\)) e^2x dx
हल:

प्रश्न 9.
समाकलन \(\int_{1 / 3}^1\) \(\frac{\left(x-x^3\right)^{1 / 3}}{x^4}\) dx का मान है:
(A) 6
(B) 0
(C) 3
(D) 4
हल:

प्रश्न 10.
यदि f(x) = \(\int_0^x\) t sin t dt, तब f(x) है:
(A) cotx + x sinx
(B) x sinx
(C) x cos x
(D) sin x + x cos x
हल:
\(\int_0^x\) t sin t dt में t को प्रथम फलन तथा sin t को द्वितीय फलन मानकर समाकलन करने पर,
f(x) = \(\int_0^x\) t sin t dt
= ∫[t – (-cost)]^x ₀ – \(\int_0^x\)1(-cost)dt
= x(-cosx) – 0 × (-cos0) + [sint]^x ₀
= -xcosx + sinx
∴ f(x) = -1.cosx – x(-sinx) + cosx
= – cosx + xsinx + cosx = xsinx
अतः विकल्प (B) सही है।