sin-5-x-dx-

Question Number 143032 by ZiYangLee last updated on 09/Jun/21

\int \sin^{- 5} x d x = ?

Answered by Olaf_Thorendsen last updated on 09/Jun/21

F (x) = \int \frac{d x}{\sin^{5} x}

Let t = \tan \frac{x}{2}

F (t) = \int \frac{1}{{(\frac{2 t}{1 + t^{2}})}^{5}} . \frac{2 d t}{1 + t^{2}}

F (t) = \frac{1}{16} \int \frac{{(1 + t^{2})}^{4}}{t^{5}} d t

F (t) = \frac{1}{16} \int \frac{1 + 4 t^{2} + 6 t^{4} + 4 t^{6} + t^{8}}{t^{5}} d t

F (t) = \frac{1}{16} \int (\frac{1}{t^{5}} + \frac{4}{t^{3}} + \frac{6}{t} + 4 t + t^{3}) d t

F (t) = \frac{1}{16} (- \frac{1}{4 t^{4}} - \frac{2}{t^{2}} + 6 \ln t + 2 t^{2} + \frac{t^{4}}{4}) + C

F (x) = - \frac{1}{{64 \tan}^{4} \frac{x}{2}} - \frac{1}{{8 \tan}^{2} \frac{x}{2}} +

\frac{3}{8} \ln (\tan \frac{x}{2}) + \frac{1}{8} \tan^{2} \frac{x}{2} + \frac{\tan^{4} \frac{x}{2}}{64} + C

F (x) = \frac{1}{8} \ln (\tan \frac{x}{2}) - (\frac{1}{{4 \sin}^{4} x} + \frac{3}{{8 \sin}^{2} x}) \cos x + C