Indefinite Integration 1 Question 63
64. 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:
- (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) $\int _1^{2} \frac{d x}{x \sqrt{x^{2}-1}}=\left[\sec ^{-1} x\right] _1^{2}=\frac{\pi}{3}-0=\frac{\pi}{3}$
