Q62 / 170 · Manipal (MET) 2020 · Mathematics
If $\int \sin \left\{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right\} d x$
$=A \sin ^{-1} x+B x \sqrt{1-x^2}+C$
then $A+B$ is equal to
- A$10$
- B$\frac{1}{2}$
- Correct answerC$1$
- D$-\frac{1}{2}$
Solution
We have,
$\int \sin \left\{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right\} d x=A \sin ^{-1} x+B x \sqrt{1-x^2}+C$
Let $\quad I=\int \sin \left\{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right\} d x$
Put $x=\cos 2 \theta \Rightarrow d x=-2 \sin 2 \theta d \theta$
$\therefore I=-\int \sin \left\{2 \tan ^{-1} \sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}\right\} 2 \sin 2 \theta d \theta$
$=-\int \sin \left\{2 \tan ^{-1} \sqrt{\frac{2 \sin ^2 \theta}{2 \cos ^2 \theta}}\right\} 2 \sin 2 \theta d \theta$
$=-\int \sin \left\{2 \tan ^{-1} \sqrt{\tan ^2 \theta}\right\} 2 \sin 2 \theta d \theta$
$=-\int \sin \left\{2 \tan ^{-1} \tan \theta\right\} 2 \sin 2 \theta d \theta$
$=-\int[\sin (2 \theta)] 2 \sin 2 \theta d \theta$
$=-\int 2 \sin ^2 2 \theta d \theta=-\int(1-\cos 4 \theta) d \theta$
$=-\left[\theta-\frac{\sin 4 \theta}{4}\right]+C$
$=\left[-\frac{1}{2} \cos ^{-1} x+\frac{1}{4} \times 2 \sin 2 \theta \cos 2 \theta\right]+C$
$=\left[-\frac{1}{2}\left(\frac{\pi}{2}-\sin ^{-1} x\right)+\frac{1}{2} \sqrt{1-x^2} \times x\right]+C$
$=\frac{1}{2} \sin ^{-1} x+\frac{x}{2} \sqrt{1-x^2}+\left(C-\frac{\pi}{4}\right)$
But $I=A \sin ^{-1} x+B x \sqrt{1-x^2}+C$
$\therefore \quad A=\frac{1}{2}$ and $B=\frac{1}{2}$
Hence, $\quad A+B=\frac{1}{2}+\frac{1}{2}=1$
Hence, the correct answer is option: 'C'.