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CGPET 2014 Solved Paper

Section: Mathematics

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Question : 119 of 150

Marks: +1, -0

If

\int \sin \left\{ 2 \tan^{-1} \sqrt{ \frac{1-x}{1+x} } \right\} dx

= A \sin^{-1} x + B x \sqrt{1-x^2} + C,

A+B

10
$\frac{1}{2}$
1
$-\frac{1}{2}$

Solution:

We have,

\int \sin \left\{ 2 \tan^{-1} \sqrt{ \frac{1-x}{1+x} } \right\} dx = A \sin^{-1} x

+ B x \sqrt{1-x^2} + C

\text{ Let } I = \int \sin \left\{ 2 \tan^{-1} \sqrt{ \frac{1-x}{1+x} } \right\} dx

= - \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 \left[ \sin (2\theta) \right] 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 A = \frac{1}{2}

and

B = \frac{1}{2}

Hence,

A+B = \frac{1}{2} + \frac{1}{2} = 1

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