How do you evaluate #\int \cos 3x \sin ^ { 2} x d x#?

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2 Answers
Andrea S.
Mar 5, 2018

#int cos3xsin^2xdx = sin^3x/3-(4sin^5x)/5+C#

Explanation:

Use the trigonometric formulas:

#cos 3x = cos(2x+x) = cos2xcosx-sin2xsinx#

#cos 3x = (cos^2x-sin^2x) cosx-2cosxsin^2x#

#cos 3x = cos^3x-3cosxsin^2x#

so:

#cos3xsin^2x = cos^3xsin^2x-3cosxsin^4x#

#cos3xsin^2x = cos^3x(1-cos^2x)-3cosxsin^4x#

#cos3xsin^2x = cos^3x- cos^5x-3cosxsin^4x#

Solve the integral separately:

#int cos^3xdx = int cosx(1-sin^2x)dx#

#int cos^3xdx = int cosxdx - int sin^2xcosxdx#

#int cos^3xdx = sinx - int sin^2xd(sinx)#

#int cos^3xdx = sinx - sin^3x/3+c_1#

then:

#int cos^5x = int cosx(1-sin^2x)^2dx#

#int cos^5x = int (1-2sin^2x+sin^4x)d(sinx)#

#int cos^5x = sinx-(2sin^3x)/3+sin^5x/5 +c_2#

and finally:

#int cosxsin^4xdx = int sin^4xd(sinx) = sin^5x/5+c_3#

Putting the partial solutions together:

#int cos3xsin^2xdx = sinx - sin^3x/3-sinx+(2sin^3x)/3-sin^5x/5-(3sin^5x)/5+C#

and simplifying:

#int cos3xsin^2xdx = sin^3x/3-(4sin^5x)/5+C#

maganbhai P.
Mar 5, 2018

#I=sin^3x/3-(4sin^5x)/5+C#

Explanation:

#color(red)(int[f(x)]^nd/(dx)f(x)dx=[f(x)]^(n+1)/(n+1)+C)#

#I=int[4cos^3x-3cosx]sin^2xdx#
#=int[4cos^3x*sin^2x-3cosx*sin^2x]dx##=int[4cosxcos^2xsin^2x-3sin^2xcosx]dx##=int[4cosx(1-sin^2x)sin^2x-3sin^2xcosx]dx##=int[4sin^2xcosx-4sin^4xcosx-3sin^2xcosx]dx##=int[sin^2xcosx-4sin^4xcosx]dx##=int(sinx)^2*d/(dx)(sinx)dx-4int(sinx)^4*d/(dx)(sinx)dx##=(sinx)^3/3-4*(sinx)^5/5+C#
#=sin^3x/3-(4sin^5x)/5+C#

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