Question

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

• A. $\ell n2$
• B. $\frac{1}{2}\ell 2$
• C. $\frac{1}{2}$
• D. $\frac{1}{\ell n2}$

Answer 1

The correct option is D $\frac{1}{\ell n2}$

Consider the following integral.

$\underset{}{\overset{}{\int }}\frac{2^x}{\sqrt{1-4^x}}dx$

Solution:

Let and differentiate w.r.t “x” we get.

$t=2^x$

$\frac{dt}{dx}=2^x\ln 2$

$\underset{}{\overset{}{\int }}\frac{2^x}{\sqrt{1-4^x}}dx......\left(1\right)$

Put the value $$dx$$ of in eq . (1) we get.

$=\underset{}{\overset{}{\int }}\frac{2^x}{\sqrt{1-4^x}}dx......\left(1\right)$

$=\underset{}{\overset{}{\int }}\frac{2^x}{2^x\ln 2\sqrt{1-{\left(t^2\right)}^x}}dx$

$=\underset{}{\overset{}{\int }}\frac{1}{\ln 2\sqrt{1-{\left(t^2\right)}^x}}dx$

$=\frac{1}{\ln 2}{\sin }^{-1}x+C$

Hence, the value of $[K=\frac{1}{\ln 2}]$

Hence, this is the correct answer .