Indefinite Integration 1 Question 19
20. The integral $\int _{\pi / 4}^{\pi / 2}(2 \operatorname{cosec} x)^{17} d x$ is equal to
(a) $\int _0^{\log (1+\sqrt{2})} 2\left(e^{u}+e^{-u}\right)^{16} d u$
(2014 Adv.)
(b) $\int _0^{\log (1+\sqrt{2})}\left(e^{u}+e^{-u}\right)^{17} d u$
(c) $\int _0^{\log (1+\sqrt{2})}\left(e^{u}-e^{-u}\right)^{17} d u$
(d) $\int _0^{\log (1+\sqrt{2})} 2\left(e^{u}-e^{-u}\right)^{16} d u$
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Solution:
- PLAN This type of question can be done using appropriate substitution.
Given, $I=\int _{\pi / 4}^{\pi / 2}(2 \operatorname{cosec} x)^{17} d x$
$$ =\int _{\pi / 4}^{\pi / 2} \frac{2^{17}(\operatorname{cosec} x)^{16} \operatorname{cosec} x(\operatorname{cosec} x+\cot x)}{(\operatorname{cosec} x+\cot x)} d x $$
Let $\quad \operatorname{cosec} x+\cot x=t$
$\Rightarrow \quad\left(-\operatorname{cosec} x \cdot \cot x-\operatorname{cosec}^{2} x\right) d x=d t$
and $\operatorname{cosec} x-\cot x=1 / t$
$$ \begin{aligned} & \Rightarrow & 2 \operatorname{cosec} x & =t+\frac{1}{t} \\ & \therefore & I & =-\int _{\sqrt{2}+1}^{1} 2^{17} \frac{t+\frac{1}{t}}{2} \quad \frac{d t}{t} \end{aligned} $$
Let $t=e^{u} \Rightarrow d t=e^{u} d u$.
When $t=1, e^{u}=1 \Rightarrow u=0$
and when $t=\sqrt{2}+1, e^{u}=\sqrt{2}+1$
$$ \begin{aligned} \Rightarrow \quad u & =\ln (\sqrt{2}+1) \\ \Rightarrow \quad I & =-\int _{\ln (\sqrt{2}+1)}^{0} 2\left(e^{u}+e^{-u}\right)^{16} \frac{e^{u} d u}{e^{u}} \\ & =2 \int _0^{\ln (\sqrt{2}+1)}\left(e^{u}+e^{-u}\right)^{16} d u \end{aligned} $$
