Use substitution to evaluate the integral:

Question

anonymous · Answer

$$\int\limits_{0}^{\pi/2} sinxe^(cosx)dx $$

anonymous · Answer

it is e^(cosx) I'm bad with the equation tool lol

anonymous · Answer

$$\int_0^{\pi/2}\sin xe^{\cos x}\,dx$$
Setting $u=\cos x$, you have $-du=\sin x\,dx$, and so
$$\int_0^{\pi/2}\sin xe^{\cos x}\,dx=-\int_1^0e^u\,du=\int_0^1e^u\,du$$

anonymous · Answer

Then what do we do? Substitute back in?

anonymous · Answer

Or take integral. Confused

anonymous · Answer

At this point you can simply integrate. $\int e^u\,du=e^u+C$

It seems to me that the change of variables might have confused you? The first integral is with respect to $x$ over the interval $\left[0,\dfrac{\pi}{2}\right]$. When you change the variable, you also have to change the limits.

With $u=\cos x$, for $x=0$ we get $u=\cos0=1$, and for $x=\dfrac{\pi}{2}$ we get $u=\cos\dfrac{\pi}{2}=0$. The negative sign accounts for the differential, $du=-\sin x\,dx$. Then I use the property that $\displaystyle -\int_b^a=\int_a^b$

Now with
$$\int_0^1e^u\,du$$
you can simply find the antiderivative and apply the fundamental theorem of calculus.

anonymous · Answer

Thank you so much!

anonymous · Answer

You're welcome!