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Find the limit of $\int \sin(x)e^{ -x}\,dx$

I would like to know how to solve the following:
Find the limit of $$\int\sin(x)e^{ -x}dx$$
As far as I know, one should solve this problem by setting $u=\sin(x)$, then one should get $du=-\cos(x)\,dx$, then it becomes $$\int\frac{du}{\cos(x)}$$ where the integral is $\tan(x)$, then it becomes $$x-\tan(x)$$ but I don’t know how to find the limit here because of the minus sign.

A:

The limits of integration need to be established. Start by splitting the integral into two pieces:
$$\int_0^\frac{\pi}4\sin(x)\ e^{ -x}\,dx+\int_{\frac{\pi}4}^\frac{\pi}2\sin(x)\ e^{ -x}\,dx$$
The second half is clearly zero. For the first half, you need to determine the limits of integration. Notice that the derivative of $\sin(x)$ satisfies
$$\sin'(x)=-\cos(x)\qquad\text{and}\qquad \cos'(x)=\sin(x)$$
Since $\sin(x)$ goes through $-1$ at $x=\frac{\pi}4$, the definite integral is equal to
\int_0^\frac{\pi}4\ \sin(x)\ e^{ -x}\,dx=-

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