Question #99e82

Question

1757 views around the world

You can reuse this answer
Creative Commons License

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:

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