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Question Number 907 by Yugi last updated on 19/Apr/15

D e t e r m i n e t h e r e s u l t s o f t h e f o l l o w i n g .

I_{1} = \int \sqrt{\frac{a + e^{x}}{a - e^{x}}} d x

I_{2} = \int \frac{(t a n x) ∣ t a n x ∣}{1 - {t a n}^{2} x} d x

Commented by prakash jain last updated on 21/Apr/15

I_{1}

e^{x} = a \cos t

e^{x} d x = - a \sin t d t

d x = - \tan t d t

\int - \tan t \frac{1}{\tan \frac{t}{2}} d t

= - \int \frac{2}{1 - \tan^{2} t / 2} d t = - \int (\sec t + 1) d t

= - t - \ln (\sec t + \tan t) + C

where t = \cos^{- 1} \frac{e^{x}}{a}