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Evaluate: $\int\frac{dx}{(3+4\sin x)^2}$

My attempt: I have tried to express the integrand in terms of $\tan x$ and $\sec x$ but there was no use since the substitution $\tan x=z$ is of no use after that. I also tried to use Weierstrass substitution but i got a very complicated algebraic expression. Please help.

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Hint: I think the general solution for these types integrals is Tangent half-angle substitution, with $$\sin x=\dfrac{2t}{1+t^2}~~~,~~~dx=\dfrac{2}{1+t^2}\ dt$$ the integral simplifies to $$\int\frac{dx}{(3+4\sin x)^2}=\int\frac{2}{(3t^2+8t+3)^2}\ dt$$ then the squaring of denominator gives the result.

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Hint:

Integrating by parts,

$$\int\dfrac{\cos x\ dx}{\cos x(a+b\sin x)^n}=\dfrac1{\cos x}\int\dfrac{\cos x\ dx}{(a+b\sin x)^n}-\int\left(\dfrac{d(\sec x)}{dx}\int\dfrac{\cos x\ dx}{(a+b\sin x)^n}\right)dx$$

$$=\dfrac{}{b(1-n)\cos x(a+b\sin x)^{n-1}}-\int\dfrac{\sin x}{b(1-n)(1-\sin^2x)(a+b\sin x)^{n-1}}$$

Here $n=2$

Now use Partial fraction, $$\dfrac{\sin x}{(1-\sin^2x)(a+b\sin x)}=\dfrac A{1+\sin x}+\dfrac B{1-\sin x}+\dfrac C{a+b\sin x}$$

and Weierstrass substitution in the last integral as the first two are elementary.

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