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.

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

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