$\int \frac{1}{\cos ^3 x \sqrt{\sin 2 x}} \mathrm{~d} x=$
$\int \frac{1}{\cos ^3 x \sqrt{\sin 2 x}} \mathrm{~d} x=$
- $\sqrt{2}\left(\sqrt{\tan x}+\frac{1}{5}(\tan x)^{\frac{5}{2}}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
- $\left(\sqrt{\tan x}+\frac{2}{5}(\tan x)^{\frac{5}{2}}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
- $\frac{1}{\sqrt{2}}\left(\sqrt{\tan x}+\frac{2}{5}(\tan x)^{\frac{5}{2}}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
- $2\left(\sqrt{\tan x}+\frac{1}{5}(\tan x)^{\frac{5}{2}}\right)+c$, where $c$ is a constant of integration.
Solution
$\text { Let } \begin{aligned}
\mathrm{I} & =\int \frac{1}{\cos ^3 x \sqrt{\sin 2 x}} \mathrm{~d} x \\
& =\frac{1}{\sqrt{2}} \int \frac{\sec ^3 x}{\sqrt{\sin x \cos x}} \mathrm{~d} \\
& =\frac{1}{\sqrt{2}} \int \frac{\sec ^4 x}{\sqrt{\tan x}} \mathrm{~d} x
\end{aligned}$
Let $\tan x=\mathrm{t} \Rightarrow \sec ^2 x=\mathrm{dt}$
$\begin{aligned}
\therefore \quad I & =\frac{1}{\sqrt{2}} \int \frac{1+t^2}{\sqrt{t}} d t \\
& =\frac{1}{\sqrt{2}} \int t^{-\frac{1}{2}} d t+\frac{1}{\sqrt{2}} \int t^{\frac{3}{2}} d t \\
& =\frac{1}{\sqrt{2}}\left(\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right)+\frac{1}{\sqrt{2}}\left(\frac{t^{\frac{5}{2}}}{\frac{5}{2}}\right)+\mathrm{c} \\
& =\sqrt{2}\left(\sqrt{\tan x}+\frac{1}{5}(\tan x)^{\frac{5}{2}}\right)+\mathrm{c}
\end{aligned}$
Asked in: MHT CET 2023 (12 May Shift 2)
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