Question

If $\int \frac{2^x}{\sqrt{1-4^x}}dx=K{\sin }^{-1}\left(2^x\right)+C$, then $K$ is equal to

Answer 1

$\int \frac{2^x}{\sqrt{1-4^x}}dx=k{\sin }^{-1}\left(2x\right)+C$

Differentiating both to sides wrt x

$\Rightarrow \frac{2^x}{\sqrt{1-4^x}}=kx\frac{1}{\sqrt{1-(2^x{)}^2}}\times 2^x\ell n^2$

$\Rightarrow \frac{2^x}{\sqrt{1-4^x}}=kx\frac{1}{\sqrt{1-2^{2x}}}\times 2^x\ell n^2$

$\therefore k=\frac{1}{\ell n^2}$