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January 31, 2015

January 31, 2015

Posted by **emath** on Tuesday, February 23, 2010 at 10:10pm.

I'm having problems just even integrating it...

I tried breaking it up

(sin^(2)x)^(2)(cos^(2)x)^(6)(sinx)dx

then doing u-sub with u=cosx and du= -sinx

but then I get stuck...

the answer is 8/3315 just don't know how to get that answer!

- Math -
**MathMate**, Wednesday, February 24, 2010 at 8:18pmWhen you have products of powers of sin and cos, look to see if one of the has odd power and the other one is even (as in this case).

This being the case, you can use reduce the odd power to even by combining it with the dx, and reduce the rest (all even power) to powers of either only sine or only cosine.

I=∫sin^5(x)cos^12(x)dx

=∫sin^4(x)cos^12(x) sin(x)dx

=∫ (1-sin²(x))²cos^12(x) sin(x)dx

Use substitution u=cos(x), du=-sin(x)dx

I=-∫(1-u²)²u^12du

=-∫(u^12-2u^14+u^16)du

=-(u^13/13-2u^15/15+u^17/17)] from u=cos(0) to u=cos(π/2)

= 8/3315

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