Where have I gone wrong in this integral?

Question

The integral in question is $$I= \int{\frac{25}{(3 \cos x+4 \sin x)^2}}dx$$

Answer 1

Hint one: $3 \cos{x} + 4 \sin{x} = 5 \cos{(x-\delta)}$.

Hint two: $$\int du \, \sec^2{u} = \tan{u} + C$$

Answer 2

You wish to determine $I = \int \frac{25}{(3 \cos x + 4 \sin x)^2} dx$, then (as you have done) write the integrand as $(\frac{5}{3 \cos x + 4 \sin x})^2$. Then this is equal to $\frac{1}{\sin^2 (x + \alpha)}$, which you have shown. So, finally you have to determine $I = \int \frac{1}{\sin^2 (x + \alpha)} dx$. Introduce the variable $y = x + \alpha$ so that the desired integral is $I = \int \frac{1}{\sin^2 (y)} dy$. This is not a particular straight forward integral to determine. Look at Evaluate $\int\frac{d\theta}{1+x\sin^2(\theta)}$ for the approach to solve this.

Answer 3

Beside the good hint given by Ron Gordon, it seems that the tangent half-angle substitution works quite well $$t=\tan(\frac x2)\qquad \cos(x)=\frac{1-t^2}{1+t^2}\qquad \sin(x)=\frac{2t}{1+t^2}\qquad dx=\frac{2\,dt}{1+t^2}$$ leads, after simplifcation and partial fraction decomposition, to $$\int \frac{25}{(3\cos(x)+4\sin(x))^2} dx=\int\frac{50 \left(t^2+1\right)}{\left(-3 t^2+8 t+3\right)^2}\,dt=5 \int\left(\frac{1}{(3 t+1)^2}+\frac{1}{(t-3)^2}\right)\,dt$$ which seems to be practicable.