Got stuck trying to do the integration of $ \int \frac{\ e^x (\cos x - \sin x)}{\sin^2 x} dx$

Question

$\int \frac{\ e^x (\cos x - \sin x)}{\sin^2 x} dx$

Tried to do multiply and got $ \int \left(\frac{\ e^x \cos x}{\sin^2 x} \right)$ - $ \int \left(\frac{\ e^x}{\sin x}\right)$, but still stuck on that.

Could you help me, please?

Answer 1

Apply integration by parts (differentiating $e^x$ and integrating $\csc x\cot x$), \begin{align} \int e^x\csc x\cot xdx-\int e^x\csc xdx&=\left[e^x(-\csc x)-\int e^x(-\csc x)dx\right]-\int e^x\csc xdx+C\\ &=-e^x\csc x+C \end{align} where $C$ is the constant of integration.

Answer 2

The integrand looks very much like a derivative that we could get via the quotient rule. Recall that the quotient rule states: $$\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{f'(x)g(x)-f(x)g'(x)}{[f(x)]^2}$$ After a bit of thought we may be able to see that the numerator is equal to $$e^x(\frac{d}{dx}\sin(x)-\sin x)=-(\sin x\frac{d}{dx}e^x-e^x\frac{d}{dx}\sin(x))$$ and so our integral is equal to $$-\frac{e^x}{\sin x}+C$$

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I hope that helps. If you have any questions please don't hesitate to ask. Much of the credit for this answer goes to Alann Rosas as a result of his recent answer here .