integration by parts of sin^2 of x

**Frostking** · October 29th 2008, 10:20 PM

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

**mr fantastic** · October 29th 2008, 10:34 PM

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.

**Frostking** · October 29th 2008, 11:39 PM

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?????

**mr fantastic** · October 30th 2008, 02:53 AM

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?????

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

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

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

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

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

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