Question

integration of (sinx)^3/(cosx)^4

Answer 1

you can see your answer

Answer 2

$Let , \ \ I = \int {\frac{\sin^{3}x}{\cos^{4}x}} \, dx \\ \\ = \int {\frac{\sin^{2}x.\sin x}{\cos^{4}x}} \, dx \\ \\ = \int {\frac{(1 - \cos^{2} x).\sin x}{\cos^{4}x}} \, dx \\ let \ \cos x = t \\ \\ \cos x \ dx = dt \\ \\ So, \\ \\ I = \int { \frac{( 1 - t^{2})}{t^{4}} \ dx \\ \\ = \int {t^{-4} \ dx - \int {t^{-2} \ dx \\ \\ = \frac{-1}{3t^{3}} + \frac{1}{t} + c \\ \\ \\ \\ = \frac{-1}{3\cos^{3}x} + \frac{1}{\cos x} + c \ \ \textbf{Ans.}$