Vtyjkyuk

Evaluate: $intfracdx(3+4sin x)^2$

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.

Hint: I think the general solution for these types integrals is Tangent half-angle substitution, with
$$sin x=dfrac2t1+t^2~~~,~~~dx=dfrac21+t^2 dt$$
the integral simplifies to
$$intfracdx(3+4sin x)^2=intfrac2(3t^2+8t+3)^2 dt$$
then the squaring of denominator gives the result.

Hint:

Integrating by parts,

$$intdfraccos x dxcos x(a+bsin x)^n=dfrac1cos xintdfraccos x dx(a+bsin x)^n-intleft(dfracd(sec x)dxintdfraccos x dx(a+bsin x)^nright)dx$$

$$=dfracb(1-n)cos x(a+bsin x)^n-1-intdfracsin xb(1-n)(1-sin^2x)(a+bsin x)^n-1$$

Here $n=2$

Now use Partial fraction, $$dfracsin x(1-sin^2x)(a+bsin x)=dfrac A1+sin x+dfrac B1-sin x+dfrac Ca+bsin x$$

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