Question #99e82

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Answer 1 · 2015-01-31T02:50:41

If you have #intsin^2(x)dx# you can integrate by parts after rearranging your integrand:

#intsin^2(x)dx=intsin(x)sin(x)dx=#

by parts where you have:

#intf(x)*g(x)dx=F(x)*g(x)-intF(x)*g'(x)dx#

Where:

#F(x)=intf(x)dx#
#g'(x)# is the derivative of #g(x)#

#=sin(x)(-cos(x))-int(-cos(x))cos(x)dx=#

#=sin(x)(-cos(x))+intcos^2(x)dx=#

but: #cos^2(x)=1-sin^2(x)# and your integral becomes:

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

#=sin(x)(-cos(x))+int1dx-intsin^2(x)dx#

So basically you can write:

#intsin^2(x)dx=sin(x)(-cos(x))+int1dx-intsin^2(x)dx#

Taking the integral with #sin^2(x)# from the right to the left (as in a normal equation) yuo get:

#2intsin^2(x)dx=sin(x)(-cos(x))+int1dx#

and finally:

#intsin^2(x)dx=(sin(x)(-cos(x))+x)/2+c#