Question Bank [2022] — Maths STD 12 Science — Question
Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]2 Marks
Question
Evaluate: $\int_0^{\frac{\pi}{4}} \frac{\tan ^3 x}{1+\cos 2 x} d x$
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Answer
$ \text { Let } I =\int_0^{\frac{\pi}{4}} \frac{\tan ^3 x}{1+\cos 2 x} d x$
$=\int_0^{\frac{\pi}{4}} \frac{\frac{\sin ^3 x}{\cos ^3 x}}{2 \cos ^2 x} d x$
$=\frac{1}{2} \int_0^{\frac{\pi}{4}} \frac{\sin ^2 x}{\cos ^5 x} d x$
$=\frac{1}{2} \int_0^{\frac{\pi}{4}} \frac{\left(1-\cos ^2 x\right) \sin x}{\cos ^5 x} d x $
Put $\cos x=t$
$ \therefore-\sin xdx = dt$
$\therefore \sin xdx =- dt $
When $x =0, t =1$ and when $x =\frac{\pi}{4}, t =\frac{1}{\sqrt{2}}$
$ \therefore I =\frac{1}{2} \int_1^{\frac{1}{\sqrt{2}}} \frac{-\left(1- t ^2\right)}{ t ^5} dt$
$=\frac{1}{2} \int_1^{\frac{1}{\sqrt{2}}} \frac{ t ^2-1}{ t ^5} dt$
$=\frac{1}{2} \int_1^{\frac{1}{\sqrt{2}}}\left( t ^{-3}- t ^{-5}\right) dt $
$=\frac{1}{2} \int_1^{\frac{1}{\sqrt{2}}} t ^{-3} dt -\frac{1}{2} \int_1^{\frac{1}{\sqrt{2}}} t ^{-5} dt$
$=\frac{1}{2}\left[\frac{ t ^{-2}}{-2}\right]_1^{\frac{1}{\sqrt{2}}}-\frac{1}{2}\left[\frac{ t ^{-4}}{-4}\right]_1^{\frac{1}{\sqrt{2}}}$
$=-\frac{1}{4}\left[\frac{1}{ t ^2}\right]_1^{\frac{1}{\sqrt{2}}}+\frac{1}{8}\left[\frac{1}{ t ^4}\right]_1^{\frac{1}{\sqrt{2}}}$
$=-\frac{1}{4}\left(\frac{1}{\frac{1}{2}}-1\right)+\frac{1}{8}\left(\frac{1}{\frac{1}{4}}-1\right)$
$=-\frac{1}{4}+\frac{3}{8}$
$\therefore I =\frac{1}{8}$
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