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Model Paper 2 — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 23 Marks
Question
Evaluate: $\int \frac{1}{\cos x-\sin x} d x$
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Answer

Let the given integral be,
$ I =\int \frac{1}{\cos x-\sin x} d x$
$\text { Putting } \cos x =\frac{1-\tan ^2\left(\frac{x}{2}\right)}{1+\tan ^2\left(\frac{x}{2}\right)} \text { and } \sin x =\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}$
$\Rightarrow I=\int \frac{d x}{\frac{1-\tan ^2\left(\frac{x}{2}\right)}{1+\tan ^2\left(\frac{x}{2}\right)}-\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}$
$=\int \frac{\sec ^2\left(\frac{x}{2}\right) d x}{1-\tan ^2\left(\frac{x}{2}\right)-2 \tan \left(\frac{x}{2}\right)}$
$\text { Let } \tan \left(\frac{x}{2}\right)= t$
$\Rightarrow \frac{1}{2} \sec ^2\left(\frac{x}{2}\right) dx = dt$
$\sec ^2\left(\frac{x}{2}\right) dx =2 dt $
$\therefore I=\int \frac{2 d t}{1-t^2-2 t}$
$=\int \frac{-2 d t}{t^2+2 t-1}$
$=\int \frac{-2 d t}{t^2+2 t+1-2}$
$=-\int \frac{2 d t}{(t+1)^2-(\sqrt{2})^2}$
$=\int \frac{2 d t}{(\sqrt{2})^2-(t-1)^2}$
$=\frac{2}{2 \sqrt{2}} \ln \left|\frac{\sqrt{2}+t+1}{\sqrt{2}-t-1}\right|+C$
$=\frac{1}{\sqrt{2}} \ln \left|\frac{\sqrt{2}+\tan \frac{x}{2}+1}{\sqrt{2}-\tan \frac{x}{2}-1}\right|+C$

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