Struggling with Integrals? Don't Panic, Here's Some Tips!

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Homework Statement



[itex]\int sin^{2}x dx[/itex]

Homework Equations





The Attempt at a Solution



I don't see why. Clearly it is designed for me to shoot myself before finishing.

u = (sinx)^2
du = 2sinxcosx dx, magically equal to sin2x dx
v = x
dv = dx

(-xcos2x/2) + (sin2x/4)

=

(-2xcos2x+sin2x)/(4) + C = wrong
 
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Firstly, let me apologize for my attitude. Secondly, thanks for the tip, I'll work it again and post back.
 
[itex]\int sin^{2}x dx[/itex]

[itex]u = sin^{2}x[/itex]
[itex]du = sin2x dx[/itex]
[itex]v = x[/itex]
[itex]dv = dx[/itex]

[itex]xsin^{2}x - \int xsin2x dx[/itex]

[itex]u = x[/itex]
[itex]du = dx[/itex]
[itex]v = -\frac{cos2x}{2}[/itex]
[itex]dv = sin2x dx[/itex]

[itex]xsin^{2}x - [-\frac{xcos2x}{2} - \int -\frac{cos2x}{2} dx ][/itex]
[itex]xsin^{2}x - [-\frac{xcos2x}{2} + \frac{sin2x}{4}][/itex]

[itex]xsin^{2}x + \frac{xcos2x}{2} - \frac{sin2x}{4}[/itex]

[itex]\frac{2xsin^{2}x + xcos2x}{2} - \frac{sin2x}{4}[/itex]


[itex]\frac{2xsin^{2}x + x(cos^{2}x - sin^{2}x)}{2} - \frac{sin2x}{4}[/itex]


[itex]\frac{2xsin^{2}x + xcos^{2}x - xsin^{2}x}{2} - \frac{sin2x}{4}[/itex]


[itex]\frac{xsin^{2}x + xcos^{2}x}{2} - \frac{sin2x}{4}[/itex]


[itex]\frac{x(sin^{2}x + cos^{2}x)}{2} - \frac{sin2x}{4}[/itex]


[itex]\frac{x}{2} - \frac{sin2x}{4}[/itex]
 
There's a much shorter way of doing this, following Sammy's hint.

sin2(x) = 1/2 * (1 - cos(2x))

This will give you an integral that you can do without having to resort to integration by parts.
 
1MileCrash said:
Firstly, let me apologize for my attitude.

th_smiley-bangheadonwall.gif


Welcome to the wonderful world of integrals! There are times when I want to kick small, defenseless animals because of this stuff. I usually listen to some thrash metal while studying math. I don't know why, but it helps!

Try to remember as many identities as you possibly can. It makes life a lot easier.