# Thread: Finding a definite integral using a given substitution.

1. ## Finding a definite integral using a given substitution.

Hi, I have been given this question and can only get so far, I'm having trouble with the last part. Any help is good thanks

2. First of all, when you let $\displaystyle u = \sin{x} \implies du = \cos{x}\,dx$, you also need to change the terminals.

$\displaystyle u(0) = 0$ and $\displaystyle u\left(\frac{\pi}{2}\right) = 1$.

So $\displaystyle \int_0^{\frac{\pi}{2}}{2\cos{x}\,e^{\sin{x}}\,dx} = \int_0^1{2e^u\,du} = \left[2e^u\right]_0^1$

Go from here.

3. Thanks Prove It.

I don't understand how $\displaystyle u(0) = 0$ and $\displaystyle u\left(\frac{\pi}{2}\right) = 1$.

What does $\displaystyle u\left(\frac{\pi}{2}\right) = 1$ mean? Do you substitute $(\frac{\pi}{2}\right)$ somewhere?

4. Originally Posted by dix
Thanks Prove It.

I don't understand how $\displaystyle u(0) = 0$ and $\displaystyle u\left(\frac{\pi}{2}\right) = 1$.

What does $\displaystyle u\left(\frac{\pi}{2}\right) = 1$ mean? Do you substitute $(\frac{\pi}{2}\right)$ somewhere?
Yes, note that $\displaystyle u(0) = 0$ is the same as writing when x =0 (ie the zero in the bracket) then u = 0 because we know that u = sin x. Same idea for the upper limit.

5. Oh right, thanks. I had my calculator in degrees instead of radians so it came out different.

I'm supposed to end up with 2(e-1), could you tell me where i've went wrong here?

6. You are evaluating $\displaystyle \left[2e^u\right]_0^1$, not $\displaystyle \left[2e^{\sin{x}}\right]_0^1$. The whole point in changing your terminals when you make the substitution is so that you DON'T have to convert the integral back to a function of $\displaystyle x$.

7. Oh right. Thanks again.