What is #int sin^3x/cos^6x dx#?

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Answer 1 · 2015-11-23T03:32:40

#int frac{sin^3x}{cos^6x} dx = -frac{1}{3cos^3x} + frac{1}{5cos^5x} + c#,

where #c# is the constant of integration.

Use the substitution #u=cosx#.

#frac{du}{dx} = -sinx#

#int frac{sin^3x}{cos^6x} dx = int frac{cos^2x-1}{cos^6x} (-sinx) dx#

#= int frac{u^2-1}{u^6} frac{du}{dx} dx#

#= int (u^{-4} - u^{-6}) du#

#= frac{u^{-3}}{-3} - frac{u^{-5}}{-5} + c#,

where #c# is the constant of integration.

#= -frac{1}{3cos^3x} + frac{1}{5cos^5x} + c#