≤ft( {a+x/a-x} + {a-x/a+x} ) dx is equal to

Correct Answer is : 2asin⁡−1(xa)+C2a ^{-1}≤ft(x/a) + C2asin−1(ax)+C

Correct Answer is : 2asin⁡−1(xa)+C2a ^{-1}≤ft(x/a) + C2asin−1(ax)+C

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AP EAMCET Engineering 2011 Solved Paper

Section: Mathematics

Question : 91 of 160

Marks: +1, -0

\int \left( \sqrt{\frac{a+x}{a-x}} + \sqrt{\frac{a-x}{a+x}} \right) dx

is equal to

Solution:

Let

I=\int \left( \sqrt{\frac{a+x}{a-x}} + \sqrt{\frac{a-x}{a+x}} \right) dx

Put

x=a \cos 2\theta \Rightarrow dx=-2a \sin 2\theta d\theta

\therefore = -\int \left( \sqrt{\frac{a+a\cos 2\theta}{a-a\cos 2\theta}} + \sqrt{\frac{a-a\cos 2\theta}{a+a\cos 2\theta}} \right)

\times 2a \sin 2\theta d\theta

=-\int \left( \sqrt{\frac{2\cos^2\theta}{2\sin^2\theta}} + \sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}} \right)

2a \sin 2\theta d\theta

=-2a \int \left( \frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta} \right) \sin 2\theta d\theta

=-2a \int \frac{1}{\sin\theta \cos\theta} \sin 2\theta d\theta

=-4a \int d\theta = -4a \theta + C_1

=-2a \cos^{-1}\left(\frac{x}{a}\right) + C_1

=-2a \left[ \frac{\pi}{2} - \sin^{-1}\left(\frac{x}{a}\right) \right] + C_1

=2a \sin^{-1}\left(\frac{x}{a}\right) + C_1 - \pi a

=2a \sin^{-1}\left(\frac{x}{a}\right) + C

, where

C = C_1 - \pi a

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