First slide

Methods of integration

Question

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

Easy

A

$\log 2$
B

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

$\frac{1}{2}$
D

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

Solution

$\begin{array}{l} Let I = \int \frac{2^{x}}{\sqrt{1 - 4^{x}}} d x = \frac{1}{\log 2} \int \frac{1}{\sqrt{1 - t^{2}}} d t \\ Putting 2^{x} = t, 2^{x} \log 2 d x = d t \\ I = \frac{1}{\log 2} \sin^{- 1} (\frac{t}{1}) + C = \frac{1}{\log 2} \sin^{- 1} (2^{x}) + C \\ ∴ K = \frac{1}{\log 2} \end{array}$

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