evaluate integral of sin^2 x cos^3 x dx

Question

,use formula

Answer 1

Given Int sin^3(x) cos^2(x) dx
Int sin^2(x) cos^2(x) sin(x) dx
Int [1 - cos^2(x)] cos^2(x) sin(x) dx

Now, let u = cos(x), du = -sin(x) dx.
Int (1 - u^2) u^2 (-du)
Int (u^4 - u^2) du
(1/5)u^5 - (1/3)u^3 + c
(1/5)cos^5(x) - (1/3)cos^3(x)+C

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Answer 2

Since sin²x=1-cos²x, the given function can be rewritten as follows:

∫sin²xcos³xdx=∫(1-cos²x)cos³xdx=∫cos³xdx - ∫cos^5(x)dx

From trig identies such as, 4cos³x=cos3x-3cosx, and 16cos^5(x)=cos5x+5cos3x+10cosx, the function above can be rewritten as follows:

∫sin²xcos³x=(1/4)∫(cos3x+3cosx)dx - (1/16)∫(cos5x+5cos3x+10cosx)dx

From anti-derivative rules such as, ∫cos(ax+b)dx=(1/a)sin(ax+b)+C, and ∫cosxdx=sinx+C, the function above can be rewritten as follows:

∫sin²xcos³xdx=(1/4)((1/3)sin3x+3sinx+C) - (1/16)((1/5)sin5x+(5/3)sin3x+10sinx+C)

                     =(1/16)(-(1/5)sin5x-(1/3)sin3x+2sinx)+C=(-1/16)((1/5)sin5x+(1/3)sin3x-2sinx)+C

Therefore, the answer is: ∫sin²xcos³xdx=(-1/16)((1/5)sin5x+(1/3)sin3x-2sinx) + C