indefinite integral

\hspace{-16}\int\frac{x^2}{\left(1+\sin 2x\right)}dx

1 Answers

30
Ashish Kothari ·

\int \frac{x^2}{1+\sin2x}dx=\int \frac{x^2}{\left( \sin x+\cos x\right)^2}dx=\int \frac{x^2}{(\sqrt{2}\sin (x+\frac{\pi}{4}))^2}dx

=\frac{1}{2}\int \frac{x^2}{\sin^2 (x+\frac{\pi}{4})}dx=\frac{1}{2}\int x^2\csc^2(x+\frac{\pi}{4}) dx=\frac{1}{2}\int \left(t-\frac{\pi}{4}\right)^2\csc^2 t .dt

=\frac{1}{2}\left[ \int t^2\csc^2 t.dt+\frac{\pi^2}{16}\int\csc^2 t.dt -\frac{\pi}{2}\int t\csc^2 t.dt \right]

=\frac{1}{2}\left[ \left( \int t^2\csc^2 t.dt\right)-\frac{\pi^2}{16}\cot t +\frac{\pi}{2}\left( t\cot t-ln\left|\sin t \right|\right)]

How to integrate the only remaining integral? [7]

Your Answer

Close [X]