JEE MainMathematicsModerateIndefinite integration

If C is arbitrary constant
then $\int\frac{a^{x}}{\sqrt{1 - a^{2x}}}dx =$

Show answer and solution

Let

I = \int\frac{a^{x}}{\sqrt{1 - a^{2x}}}dx.

Put

a^{x} = t

So that

a^{x}\log_{e}adx = dt

dx = dt/(a^{x}\log_{e}a)

\therefore\

I = \frac{1}{\log_{e}a}\int\frac{dt}{\sqrt{1^{2} - t^{2}}}

= \frac{1}{\log_{e}a} \cdot \sin^{- 1}(t) + C = \frac{1}{\log_{e}a}\sin^{- 1}\left( a^{x} \right) + C

Source: Grass-G / M-12 Mathematics Paper 2 Nov.docx

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