Mathematics

# $\displaystyle \int \frac{\tan 2\theta d\theta }{\sqrt{\cos ^{6}\theta +\sin ^{6}\theta }}$ is equal to

$\ln\left |\displaystyle \frac{1+\sqrt{1+3\cos ^{2}2\theta }}{\cos 2\theta } \right |+c$

##### SOLUTION

$\displaystyle I= \int \frac{\tan 2\theta d\theta }{\sqrt{\cos ^{6}\theta +\sin ^{6}\theta }}$

$\displaystyle= \int \frac{\sin 2\theta d\theta }{\cos 2\theta \sqrt{1 - 3\sin ^{2}\theta \cos^{2}\theta }}$

$\displaystyle= \int \frac{\sin 2\theta d\theta }{\cos 2\theta \sqrt{1 - \dfrac{3}{4}\sin ^{2}2\theta }}$

$\displaystyle= \int \frac{2\sin 2\theta d\theta }{\cos 2\theta \sqrt{1 + 3\cos^{2}2\theta }}$
Put $\cos 2\theta = t$
$\Rightarrow -2\sin 2\theta d\theta = dt$

$\Rightarrow\displaystyle I = \int \frac{-dt}{t\sqrt{1 + 3t^{2}}}$
Put  $1 + 3t^{2} =z^2$
$\Rightarrow 6t dt = 2zdz$

So, $\displaystyle I = \int \frac{dz}{1-z^{2}}$

$\displaystyle = \frac{1}{2}\ln \left | \frac{1 + z}{1 - z} \right | + c_{1}$

$\displaystyle=\frac{1}{2}\ln \left | \frac{1 + \sqrt{1 + 3t^{2}}}{1 - \sqrt{1 + 3t^{2}}} \right | + c_{1}$

$\displaystyle = \ln \sqrt{\frac{1 + \sqrt{1 + 3t^{2}}}{1 - \sqrt{1 + 3t^{2}}}} + c_{1}$
$\displaystyle = \ln \left | \frac{1 + \sqrt{1 + 3t^{2}}}{t} \right | + c$
$\displaystyle = \ln \left | \frac{1 + \sqrt{1 + 3\cos ^{2}2\theta }}{\cos 2\theta } \right | + c$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

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