Đáp án:
Giải thích các bước giải:
$\int \dfrac{\cos ^5\left(x\right)}{1-\sin \left(x\right)}dx$
Đặt `u=sinx`
$=\int \:-\left(u+1\right)^2\left(u-1\right)du$
$=-\int \left(u+1\right)^2\left(u-1\right)du$
$=-\int \:u^3+u^2-u-1du$
$=-\left(\dfrac{u^4}{4}+\dfrac{u^3}{3}-\dfrac{u^2}{2}-u\right)$
$=-\left(\dfrac{\sin ^4\left(x\right)}{4}+\dfrac{\sin ^3\left(x\right)}{3}-\dfrac{\sin ^2\left(x\right)}{2}-\sin \left(x\right)\right)$
$=-\dfrac{1}{4}\sin ^4\left(x\right)-\dfrac{1}{3}\sin ^3\left(x\right)+\dfrac{1}{2}\sin ^2\left(x\right)+\sin \left(x\right)+C$
;
$\int \dfrac{\sin ^4\left(x\right)}{\cos ^2\left(x\right)}dx$
$=\sin ^4\left(x\right)\tan \left(x\right)-\int \:4\sin ^3\left(x\right)\cos \left(x\right)\tan \left(x\right)dx$
$=\sin ^4\left(x\right)\tan \left(x\right)-\int \:4\sin ^3\left(x\right)\cos \left(x\right)\tan \left(x\right)dx$
$=\sin ^4\left(x\right)\tan \left(x\right)-4\left(-\sin ^3\left(x\right)\cos \left(x\right)+\dfrac{3}{8}\left(x-\dfrac{1}{4}\sin \left(4x\right)\right)\right)+C$
'
$\int \left[\sin ^2\left(x\right)+\dfrac{1}{1+\cos \left(2x\right)}\right]dx$
$=\int \sin ^2\left(x\right)dx+\int \dfrac{1}{1+\cos \left(2x\right)}dx$
$=\dfrac{1}{2}\left(x-\dfrac{1}{2}\sin \left(2x\right)\right)+\dfrac{1}{2}\tan \left(x\right)$
$=\dfrac{1}{2}\left(x-\dfrac{1}{2}\sin \left(2x\right)\right)+\dfrac{1}{2}\tan \left(x\right)+C$
