1. Calculus on trigonometry... help!

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

2. Re: Calculus on trigonometry... help!

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

3. Re: Calculus on trigonometry... help!

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

4. Re: Calculus on trigonometry... help!

Hello, yugimutoshung!

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

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

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

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

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

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

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

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

5. Re: Calculus on trigonometry... help!

Beautifully done, Soroban! Thanks!