Math Help - integration by parts of sin^2 of x

1. integration by parts of sin^2 of x

I know that I can use an identity to solve this but I am suppose to use integration by parts and I can not figure out how!!! Any advice would be much appreciated. Thank you

2. Originally Posted by Frostking
I know that I can use an identity to solve this but I am suppose to use integration by parts and I can not figure out how!!! Any advice would be much appreciated. Thank you
Let $I = \int \sin^2 x \, dx$.

Let $u = \sin x$ and $dv = \sin x \, dx$:

$I = -\sin x \cos x + \int \cos^2 x \, dx = -\sin x \cos x + \int dx - I$.

Now solve for I by making it the subject.

3. integral sin^2 x by parts

I think that I must be the stupidest person alive but I can not get this.

I get sin(x)(-cos(x) - integral of (-cos(x) times cos(x) dx

which goes to sin(x)(-cos(x) + integral of cos^2(x) dx

and now I have the same type of problem all over again and I do not see how using by parts will be anything but an endless loop!!! I know I am missing the point but I feel helpless and hopeless to figure out why. Can anyone explain it to me?????

4. Originally Posted by mr fantastic
Let $I = \int \sin^2 x \, dx$.

Let $u = \sin x$ and $dv = \sin x \, dx$:

$I = -\sin x \cos x + \int \cos^2 x \, dx = -\sin x \cos x + \int dx - I$.

Now solve for I by making it the subject.
Originally Posted by Frostking
I think that I must be the stupidest person alive but I can not get this.

I get sin(x)(-cos(x) - integral of (-cos(x) times cos(x) dx

which goes to sin(x)(-cos(x) + integral of cos^2(x) dx

and now I have the same type of problem all over again and I do not see how using by parts will be anything but an endless loop!!! I know I am missing the point but I feel helpless and hopeless to figure out why. Can anyone explain it to me?????
$I = -\sin x \cos x + \int \cos^2 x \, dx$

${\color{red}= -\sin x \cos x + \int (1 - \sin^2 x) \, dx = -\sin x \cos x + \int 1 \, dx - \int \sin^2 x \, dx}$

$= -\sin x \cos x + \int dx - I$

${\color{red} \Rightarrow 2I = -\sin x \cos x + \int dx = -\sin x \cos x + x + C}$

${\color{red} \Rightarrow I = -\frac{1}{2} \sin x \cos x + \frac{x}{2} + K}$.