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The following integral    $\int_{\pi/4}^{\pi/2} (2 cosec$ $x)^{17}$ is equal t

Answer:

Explanation:

 Plan  This type of question can be done using appropriate subsitiution

   Given,   $I= \int_{\pi/4}^{\pi/2} (2 cosec$ $x)^{17}$ dx

                    $2^{17}(cosec$ $x)^{16}$cosec$x$

  $=\int_{\pi/4}^{\pi/2} \frac{(cosec x+cot x)}{(cosec x+cot x)}dx$

      Let $cosec$ $x +cot$ $x= t$

$\Rightarrow$   $(-cosec$ $x.cot$ $x-cosec^{2}$ $x)$dx= dt

 and   $cosec$ $x-cot$ x=1/t

  $\Rightarrow 2cosec$ $x=t+\frac{1}{t}$

  $\therefore I= -\int_{\sqrt{2}+1}^{1} 2^{17}\left(\frac{t+1/t}{2}\right)^{16}\frac{dt}{t}$

  Let t= e^u   $\Rightarrow$  dt = e^u du. when t=1 

 e^u  =1

$\Rightarrow$

u=0

 and when  $t=\sqrt{2}+1,e^{u}=\sqrt{2}+1$ 

$\Rightarrow$     $u= ln(\sqrt{2}+1)$

$\Rightarrow$  $I=-\int_{ln(\sqrt{2}+1)}^{0} 2(e^{u}+e^{-u})^{16}\frac{e^{u}du}{e^{u}}$

=  $2\int_{0}^{ln(\sqrt{2}+1)}(e^{u}+e^{-u})^{16}du$