Indefinite Integration 1 Question 10
11. The integral $\int _{\pi / 6}^{\pi / 4} \frac{d x}{\sin 2 x\left(\tan ^{5} x+\cot ^{5} x\right)}$ equals
(a) $\frac{1}{5} \frac{\pi}{4}-\tan ^{-1} \frac{1}{3 \sqrt{3}}$
(b) $\frac{1}{20} \tan ^{-1} \frac{1}{9 \sqrt{3}}$
(c) $\frac{1}{10} \frac{\pi}{4}-\tan ^{-1} \frac{1}{9 \sqrt{3}}$
(d) $\frac{\pi}{40}$
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Solution:
- Let $I=\int _{\pi / 6}^{\pi / 4} \frac{d x}{\sin 2 x\left(\tan ^{5} x+\cot ^{5} x\right)}$
$=\int _{\pi / 6}^{\pi / 4} \frac{\left(1+\tan ^{2} x\right) \tan ^{5} x}{2 \tan x\left(\tan ^{10} x+1\right)} d x \quad \because \sin 2 x=\frac{2 \tan x}{1+\tan ^{2} x}$
$=\frac{1}{2} \int _{\pi / 6}^{\pi / 4} \frac{\tan ^{4} x \sec ^{2} x}{\left(\tan ^{10} x+1\right)} d x$
Put $\tan ^{5} x=t\left[\because \sec ^{2} x=1+\tan ^{2} x\right]$
$\Rightarrow 5 \tan ^{4} x \sec ^{2} x d x=d t$
$$ \begin{aligned} & \begin{array}{ccc} \hline x & \frac{\pi}{6}^{\frac{\pi}{4}} \\ \hline t & \frac{1}{\sqrt{3}}^{5} & 1 \end{array} \\ & \therefore I=\frac{1}{2} \cdot \frac{1}{5} \int _{(1 / \sqrt{3})^{5}}^{1} \frac{d t}{t^{2}+1}=\frac{1}{10}\left(\tan ^{-1}(t)\right) _{(1 / \sqrt{3})^{5}}^{1} \\ & =\frac{1}{10} \tan ^{-1}(1)-\tan ^{-1} \frac{1}{9 \sqrt{3}} \\ & =\frac{1}{10} \frac{\pi}{4}-\tan ^{-1} \frac{1}{9 \sqrt{3}} \end{aligned} $$
