# Calculus on trigonometry... help!

#### yugimutoshung

"integral"{cos(2x)/[cos^2(x)*sin^2(x)]}*dx
Full-length solution for this would be really helpful!

#### Shakarri

Expand cos(2x) to get it in terms of angle x instead of 2x
Use the substitution $$\displaystyle u=cos^2x$$ and make use of $$\displaystyle cos^2x+sin^2x=1$$

1 person

#### yugimutoshung

u = cos^2(x) or u = cos (x) ?

#### Soroban

MHF Hall of Honor
Hello, yugimutoshung!

$$\displaystyle \int \frac{\cos(2x)}{\cos^2(x)\sin^2(x)}\,dx$$

Note that: .$$\displaystyle \cos^2(x)\sin^2(x) \:=\:\big[\sin(x)\cos(x)\big]^2$$

Also that: .$$\displaystyle \sin(x)\cos(x) \:=\:\tfrac{1}{2}\sin(2x)$$

The denominator becomes: .$$\displaystyle \left(\tfrac{1}{2}\sin(2x)\right)^2 \:=\:\tfrac{1}{4}\sin^2(2x)$$

The integral becomes: .$$\displaystyle \int\frac{\cos(2x)\,dx}{\frac{1}{4}\sin^2(2x)} \;=\;4\int\frac {\cos(2x)\,dx}{\sin^2(2x)}$$

Let $$\displaystyle u \,=\,\sin(2x) \quad\Rightarrow\quad du \,=\,2\cos(2x)\,dx \quad\Rightarrow\quad \cos(2x)\,dx \,=\,\tfrac{1}{2}du$$

Substitute: .$$\displaystyle 4\int\frac{\frac{1}{2}du}{u^2} \;=\;2\int u^{-2}du \;=\;2\left(\frac{u^{-1}}{-1}\right) + C \;=\;-\frac{2}{u}+C$$

Back-substitute: .$$\displaystyle -\frac{2}{\sin(2x)} + C \;=\;-2\csc(2x) + C$$

#### yugimutoshung

Beautifully done, Soroban! Thanks!