Question

Evalute the integral (sin^3 (x)/cosx)dx

Answer 1

$\int { \frac{sin ^{3} x}{cos x} } \, dx= \\ = \int { \frac{sin ^{2} x*sin x}{cos x} x} \, dx= \\ = \int { \frac{(1-sin ^{2} x)*sinx}{cos x} } \, dx =$
t = cos x,
dt = - sin x dx;
$= \int { \frac{-(1- t^{2}) }{t} } \, dt= \\ = \int { \frac{t ^{2}-1 }{t} } \, dt = \\ =\int {t} \, dt - \int { \frac{1}{t} } \, dt = t^{2}- ln t = \\ = \frac{cos ^{2} x}{2}-ln ( cos x ) + C$