Indefinite Integration Question 2
Question 2 - 30 January - Shift 2
If $\int \sqrt{\sec 2 x-1} d x=\alpha \log _{e}|\cos 2 x+\beta+\sqrt{\cos 2 x(1+\cos \frac{1}{\beta} x)}|$ + constant, then $\beta-\alpha$ is equal to ________
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Answer: 1
Solution:
Formula: Integration by Substitution, Integration of type $\int \frac{d x}{\sqrt{a x^{2}+b x+c}}$, Reciprocal Identities, Double Angle Identities
$ \begin{aligned} & \int \sqrt{\sec 2 x-1} d x=\int \sqrt{\frac{1-\cos 2 x}{\cos 2 x}} d x \\ & =\sqrt{2} \int \frac{\sin x}{\sqrt{2 \cos ^{2} x-1}} d x \end{aligned} $
Put $\cos x=t \quad \Rightarrow-\sin x d x=d t$
$=-\sqrt{2} \int \frac{dt}{\sqrt{2 t^{2}-1}}$
$=-\ln |\sqrt{2} \cos x+\sqrt{\cos 2 x}|+c$
$=-\frac{1}{2} \ln |2 \cos ^{2} x+\cos 2 x+2 \sqrt{\cos 2 x} \cdot \sqrt{2} \cos x|+$
$=-\frac{1}{2} \ln |\cos 2 x+\frac{1}{2}+\sqrt{\cos 2 x} \cdot \sqrt{1+\cos 2 x}|+c$
$\alpha = -\frac{1}{2} , \beta=\frac{1}{2}$
$\therefore \beta - \alpha = 1$
