Discussion Forum

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}  \cot^{9} x dx=$

Answer:

Explanation:

I= $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}  \cot^{9} x dx$

 $I= \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}  \frac{\cos ^{9}x}{\sin^{9}x} dx$

 put sin x=t, cos x dx=dt

 $x=\frac{\pi}{4}, t= \frac{1}{ \sqrt{2}}, x= \frac{\pi}{2},t=1$

 $\therefore$    $I= \int_{\frac{1}{\sqrt{2}}}^{1}  \frac{(1-t^{2}) ^{4}}{t^{9}}  dt$

 $I= \int_{\frac{1}{\sqrt{2}}}^{1}\left(  \frac{1-4t^{2}+6t^{4}-4t^{6}+t^{8}}{t^{9}}\right)  dt$

$I= \int_{\frac{1}{\sqrt{2}}}^{1}\left( t^{-9}-4t^{-7}+6t^{-5}-4t^{-3}+ \frac{1}{t}\right)  dt$

  $I=\left[\frac{1}{8t^{8}}-\frac{4}{6t^{6}}+\frac{6}{4t^{4}}-\frac{4}{2t^{2}}+\log t\right]^{1}_{1/\sqrt{2}}$

 I= $\left[\left(\frac{1}{8}-\frac{4}{6}+\frac{6}{4}-\frac{4}{2}+0\right)-\left(2-\frac{32}{6}-6-4-\frac{1}{2}\log 2\right)\right]$

I= $\frac{-7}{24}+\frac{1}{2} log 2$