Mathematics

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


ANSWER

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


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
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