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Vilka [71]
3 years ago
9

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

Mathematics
1 answer:
8090 [49]3 years ago
8 0
\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

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