SATHEE: Definite Integration Question 63

Definite Integration Question 63

Question 63

Match the conditions/expressions in Column I with statement in Column II.

Column I		Column II
A. $\int_{- 1}^{1} \frac{d x}{1 + x^{2}}$	P.	$\frac{1}{2} \log \frac{2}{3}$
B. $\int_{0}^{1} \frac{d x}{\sqrt{1 - x^{2}}}$	Q.	$2 \log \frac{2}{3}$
C. $\int_{2}^{3} \frac{d x}{1 - x^{2}}$	R.	$\frac{π}{3}$
D. $\int_{1}^{2} \frac{d x}{x \sqrt{x^{2} - 1}}$	S.	$\frac{π}{2}$

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Solution:

(A) Let $I = \int_{- 1}^{1} \frac{d x}{1 + x^{2}}$

Put $x = \tan θ \Rightarrow d x = \sec^{2} θ d θ$

$∴ I = 2 \int_{0}^{π / 4} d θ = \frac{π}{2}$

(B) Let $I = \int_{0}^{1} \frac{d x}{\sqrt{1 - x^{2}}}$

Put $x = \sin θ$

$\Rightarrow d x = \cos θ d θ$

$∴ I = \int_{0}^{π / 2} 1 d θ = \frac{π}{2}$

(C) $\int_{2}^{3} \frac{d x}{1 - x^{2}} = \frac{1}{2} \log {\frac{1 + x}{1 - x}}_{2}^{3}$

$= \frac{1}{2} \log \frac{4}{- 2} - \log \frac{3}{- 1} = \frac{1}{2} \log \frac{2}{3}$

(D) $\int_{1}^{2} \frac{d x}{x \sqrt{x^{2} - 1}} = {[\sec^{- 1} x]}_{1}^{2} = \frac{π}{3} - 0 = \frac{π}{3}$

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