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
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
First of all, when you let $\displaystyle \displaystyle u = \sin{x} \implies du = \cos{x}\,dx$, you also need to change the terminals.
$\displaystyle \displaystyle u(0) = 0$ and $\displaystyle \displaystyle u\left(\frac{\pi}{2}\right) = 1$.
So $\displaystyle \displaystyle \int_0^{\frac{\pi}{2}}{2\cos{x}\,e^{\sin{x}}\,dx} = \int_0^1{2e^u\,du} = \left[2e^u\right]_0^1$
Go from here.
Thanks Prove It.
I don't understand how $\displaystyle \displaystyle u(0) = 0$ and $\displaystyle \displaystyle u\left(\frac{\pi}{2}\right) = 1$.
What does $\displaystyle \displaystyle u\left(\frac{\pi}{2}\right) = 1$ mean? Do you substitute $\displaystyle (\frac{\pi}{2}\right)$ somewhere?
You are evaluating $\displaystyle \displaystyle \left[2e^u\right]_0^1$, not $\displaystyle \displaystyle \left[2e^{\sin{x}}\right]_0^1$. 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 $\displaystyle \displaystyle x$.