# Trig substitution

• Apr 30th 2011, 09:05 PM
Hardwork
Trig substitution
How can we find the integral $\displaystyle \int_{0}^{\pi}\cos{x}\sin{x}\,{dx}$ with the substitution $\displaystyle u = sinx$?

I know how to find the integral, but the limits become the same sin(0) = sin(pi) = 0.

What should I do in general when a given substitution makes upper bound = lower bound?
• Apr 30th 2011, 09:21 PM
TheChaz
Quote:

Originally Posted by Hardwork
How can we find the integral $\displaystyle \int_{0}^{\pi}\cos{x}\sin{x}\,{dx}$ with the substitution $\displaystyle u = sinx$?

I know how to find the integral, but the limits become the same sin(0) = sin(pi) = 0.

What should I do in general when a given substitution makes upper bound = lower bound?

Using a different property, you can analyse the integrand as a function with period "pi", the (signed) area of which equals zero after each period.
• Apr 30th 2011, 09:24 PM
chisigma
Quote:

Originally Posted by Hardwork
How can we find the integral $\displaystyle \int_{0}^{\pi}\cos{x}\sin{x}\,{dx}$ with the substitution $\displaystyle u = sinx$?

I know how to find the integral, but the limits become the same sin(0) = sin(pi) = 0.

What should I do in general when a given substitution makes upper bound = lower bound?

Is...

http://quicklatex.com/cache3/ql_0a43...4bcb647_l3.png

http://quicklatex.com/cache3/ql_5801...1eb2dab_l3.png

... so that, no matter which 'substitution' You do, the integral is 0...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• May 2nd 2011, 01:32 AM
Hardwork
Okay. I do see how the integral is zero. I just want to know what to do in general when a substitution makes the limits the same.