Question

If $\int \frac{2^x}{\sqrt{1-4^x}}dx=\lambda {\sin }^{-1}{2}^x+C$, then $\lambda$ equals to

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

Answer 1

Answer: (D)

$\frac{1}{\log 2}$

LHS $=I=\int \frac{2^x}{\sqrt{1-4^x}}dx=\int \frac{2^x}{\sqrt{1-{\left(2^x\right)}^2}}dx\_\_\_\left(1\right)$

Let $2^x=a\Rightarrow 2^x\log 2dx=da$ or $2^{x}dx=\frac{da}{\log 2}\_\_\_\left(2\right)$

$\Rightarrow x={\log }_{2}a$

Now from $\left(1\right)$ and $\left(2\right)$

$I=\int \frac{1}{\log 2}\frac{da}{\sqrt{1-a^2}}=\frac{1}{\log 2}{\sin }^{-1}a+c$

$=\frac{1}{\log 2}{\sin }^{-1}{2}^x+c$

So we get $\lambda =\frac{1}{\log 2}$