U-Substitution

Feb 2010
98
0
I need help with the u-substituion. The first one is....

(2cosx)/(1-cos^2x)
 

Chris L T521

MHF Hall of Fame
May 2008
2,844
2,046
Chicago, IL
I need help with the u-substituion. The first one is....

(2cosx)/(1-cos^2x)
Hint: \(\displaystyle 1-\cos^2x=\sin^2x\). Now what is the substitution?
 

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
I really don't know..
better brush up on your basic identities ...

\(\displaystyle \frac{2\cos{x}}{1 - \cos^2{x}} = \frac{2\cos{x}}{\sin^2{x}} = 2\csc{x}\cot{x}
\)
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
Notice that if \(\displaystyle u = \sin{x}\) then \(\displaystyle du = \cos{x}\,dx\).


So \(\displaystyle \int{\frac{2\cos{x}}{\sin^2{x}}\,dx} = 2\int{u^{-2}\,du}\)

\(\displaystyle = -2u^{-1} + C\)

\(\displaystyle = -\frac{2}{\sin{x}}+ C\)

\(\displaystyle = -2\csc{x} + C\).
 
Mar 2010
715
381
You can also let \(\displaystyle u = 1-\cos^2{x}\), then \(\displaystyle \dfrac{du}{dx} = -2\cos{x}\sin{x} \Rightarrow {dx} = \dfrac{du}{-2\cos{x}\sin{x}}
\) so then \(\displaystyle \int\dfrac{2\cos{x}}{1-\cos^2{x}}\;{dx} = \int\dfrac{2\cos{x}}{u\left(-2\cos{x}\sin{x}\right)}\;{du} = -\int\dfrac{1}{u\sin{x}}\;{du} \). Now since \(\displaystyle u = 1-\cos^2{x} = \sin^2{x}\), then \(\displaystyle \sin{x} = \sqrt{u}\). So we have: \(\displaystyle -\int\dfrac{1}{u\sin{x}}\;{du}= -\int\dfrac{1}{u\sqrt{u}}\;{du} = -\int\dfrac{1}{\sqrt{u^3}}\;{du} = \dfrac{-2}{\sqrt{u}}+C\) \(\displaystyle \Rightarrow \int\dfrac{2\cos{x}}{1-\cos^2{x}}\;{dx} = \dfrac{-2}{\sqrt{1-\cos^2{x}}}+C.
\)