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Find the indefinite integral: $$ \int\frac{25}{(3\cos(x)+4\sin(x))^2} dx$$

Obviously the first thing to is to expand the denominator but that doesn't help us that much. I have also tried applying the Weierstrass substitution but that lead to a black hole of algebra. I also notice the denominator has a $3$ and $4$ and the numerator has $25$ which mean a $3-4-5$ triangle might be related to this integral but I'm not to sure.

Any help is appreciated :)

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    $\begingroup$ Hint: $3\sin(x)+4 \cos(x)=5 \sin(x+\arctan(4/3))$ $\endgroup$ – tired Dec 20 '16 at 12:09
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    $\begingroup$ Furthermore $\int dx \sin(x)^{-2}=\text{cotan(x)}$ $\endgroup$ – tired Dec 20 '16 at 12:12
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Hint. One may write $$ \int \frac{25}{(3\cos(x)+4\sin(x))^2} dx=\int \frac{25}{(3+4\tan(x))^2} \frac{dx}{\cos^2 x}=\int \frac{25}{(3+4u)^2} \:du $$ with $u=\tan x$.

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    $\begingroup$ simple and elegant. $\endgroup$ – Juniven Dec 20 '16 at 12:49
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Let $\varphi$ such that $\;\begin{cases}\cos\varphi=\frac35,\\\sin\varphi=\frac45.\end{cases}$, e.g. $\;\varphi=\arctan\bigl(\frac43\bigr)$.

The integral can be rewritten as $$\int\frac{\mathrm dx}{\cos^2(x-\varphi)}=\tan(x-\varphi).$$

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