Calculus on trigonometry... help!

Apr 2013
19
0
Malaysia
"integral"{cos(2x)/[cos^2(x)*sin^2(x)]}*dx
Full-length solution for this would be really helpful!
 
Oct 2012
751
212
Ireland
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 \(\displaystyle cos^2x+sin^2x=1\)
 
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Soroban

MHF Hall of Honor
May 2006
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Lexington, MA (USA)
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: .\(\displaystyle -\frac{2}{\sin(2x)} + C \;=\;-2\csc(2x) + C\)