Giải thích các bước giải:
Ta có:
$P=\displaystyle\int\dfrac{\sin^3x}{2\cos x+1}dx$
$\to P=\displaystyle\int\dfrac{\sin^2x\cdot \sin x}{2\cos x+1}dx$
$\to P=\displaystyle\int\dfrac{(1-\cos^2x)\cdot \sin x}{2\cos x+1}dx$
Đặt $\cos x=t\to d(\cos x)=dt\to -\sin xdx=dt$
$\to P=-\displaystyle\int\dfrac{(1-t^2)}{2t+1}dt$
$\to P=-\displaystyle\int -\dfrac12t+\dfrac14+\dfrac3{4(2t+1)}dt$
$\to P=-(\dfrac14t^2+\dfrac14t+\dfrac38\ln|2t+1|)+C$
$\to P=-(\dfrac14\cos^2x+\dfrac14\cos x+\dfrac38\ln|2\cos x+1|)+C$
$\to I=\displaystyle\int^{\pi}_0\dfrac{\sin^3x}{2\cos x+1}dx=(-(\dfrac14\cos^2(\pi)+\dfrac14\cos(\pi)+\dfrac38\ln|2\cos (\pi)+1|))-(-(\dfrac14\cos^20+\dfrac14\cos0+\dfrac38\ln|2\cos0+1|))$
$\to I=\dfrac{3\ln3+4}8$