Integration help when n&m are both the same

**Pi314** · March 19th 2009, 10:20 PM

I need help with solving the integral of:
$cos^3(x)sin^3(x)dx$

I got:
$(1/4)sin^4(x)-(1/5)sin^5(x)+c$ as the answer but I don't think it's correct...

I tried to use the table of integrals I have but I couldn't find something that met that scenario so I ended up substituting u for sin(x) as well as du for cos(x) and had the following before substituting u back in, I used the trig. property of cos^2(x) = 1-sin^2(x):
Integral of $u^3(1-u^2)du$

Then:
Integral of $(u^3-u^5)du$

Thanks,

**Moo** · March 19th 2009, 11:32 PM

Hello,

Originally Posted by Pi314

I need help with solving the integral of:
$cos^3(x)sin^3(x)dx$

I got:
$(1/4)sin^4(x)-(1/5)sin^5(x)+c$ as the answer but I don't think it's correct...

I tried to use the table of integrals I have but I couldn't find something that met that scenario so I ended up substituting u for sin(x) as well as du for cos(x) and had the following before substituting u back in, I used the trig. property of cos^2(x) = 1-sin^2(x):
Integral of $u^3(1-u^2)du$

Then:
Integral of $(u^3-u^5)du$

Thanks,

Your substitution is correct

But there is a problem for the integration.

$\int u^5 ~du=\frac{u^6}{6}+c$
and you wrote $\frac{\sin^5(x)}{5}$

**Pi314** · March 20th 2009, 05:04 PM

Wow, I can't believe I didn't spot that.

Thanks.

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