JEE Main 2023 — Indefinite Integration Question with Solution

Given,

$\int \sqrt{\sec 2x-1}dx=\alpha {\log }_e\left|\cos 2x+\beta +\sqrt{\cos 2x\left(1+\cos \frac{1}{\beta }x\right)}\right|+C$

Now solving $\mathrm{L}.\mathrm{H}.\mathrm{S}$ we get,

$\int \sqrt{\sec 2x-1}dx=\int \sqrt{\frac{1-\cos 2x}{\cos 2x}}dx$

$\Rightarrow \int \sqrt{\sec 2x-1}dx=\sqrt{2}\int \frac{\sin x}{\sqrt{2{\cos }^{2}x-1}}dx$

Now let $\cos x=t \Rightarrow -\sin x dx=dt$

$\Rightarrow \int \sqrt{\sec 2x-1}dx=-\sqrt{2}\int \frac{dt}{\sqrt{2t^2-1}}$

$\Rightarrow \int \sqrt{\sec 2x-1}dx=-\ln |\sqrt{2}\cos x+\sqrt{\cos 2x}|+C$

$=-\frac{1}{2}\ln \left|2 {\cos }^{2}x+\cos 2x+2\sqrt{\cos 2x}·\sqrt{2 }\cos x\right|+C$

$=-\frac{1}{2}\ln \left|\cos 2x+\frac{1}{2}+\sqrt{\cos 2x}·\sqrt{1+\cos 2x}\right|+C$

Now on comparing with $\int \sqrt{\sec 2x-1}dx=\alpha {\log }_e\left|\cos 2x+\beta +\sqrt{\cos 2x\left(1+\cos \frac{1}{\beta }x\right)}\right|+C$

We get, $\beta =\frac{1}{2},\alpha =-\frac{1}{2}\Rightarrow \beta -\alpha =1$