Definite Integration Question 63

Question 63

  1. Match the conditions/expressions in Column I with statement in Column II.
Column I Column II
A. $\quad \int_{-1}^{1} \frac{d x}{1+x^{2}}$ 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. $\quad \int_{2}^{3} \frac{d x}{1-x^{2}}$ R. $\frac{\pi}{3}$
D. $\quad \int_{1}^{2} \frac{d x}{x \sqrt{x^{2}-1}}$ S. $\frac{\pi}{2}$
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Solution:

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

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

$$ \therefore \quad I=2 \int_{0}^{\pi / 4} d \theta=\frac{\pi}{2} $$

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

Put $\quad x=\sin \theta$

$\Rightarrow d x=\cos \theta d \theta$

$\therefore \quad I=\int_{0}^{\pi / 2} 1 d \theta=\frac{\pi}{2}$

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



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