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Integration Problem: int (cos 2x) / (1+cos 2x)^1/2 dx
- Thread starter Lizzy
- Start date
Count Iblis
Junior Member
- Joined
- Feb 13, 2007
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- 108
Re: Integration Problem
Use that cos(2x) = 2 cos^2(x) -1
Lizzy said:I need help with this integral:
The integral of (cos 2x) / (1+cos 2x)^1/2 dx
Thanks for helping out.
Use that cos(2x) = 2 cos^2(x) -1
Hello, Lizzy!
To find the integral of (cos 2x) / (1 + cos 2x)<sup>1/2</sup> dx, use (as the previous tutor suggested) the fact that cos 2x = 2 cos<sup>2</sup> - 1:
(cos 2x) / (1 + cos 2x)<sup>1/2</sup>
. . .= ( 2 cos<sup>2</sup> x - 1 ) / ( 2 cos<sup>2</sup> x )<sup>1/2</sup>
. . .= ( 2 cos<sup>2</sup> x - 1 ) / 2<sup>1/2</sup> cos x
. . .= ( 2 cos<sup>2</sup> x ) / ( 2<sup>1/2</sup> cos x )- 1 / ( 2<sup>1/2</sup>cos x )
. . .= 2<sup>1/2</sup> cos x - 2<sup>1/2</sup> sec x
integral ( 2<sup>1/2</sup> cos x - 2<sup>1/2</sup> sec x ) dx
. . .= 2<sup>1/2</sup> sin x - 2<sup>1/2</sup> ln | sec x + tan x | + C
If you have questions, please reply. Thank you! :wink:
To find the integral of (cos 2x) / (1 + cos 2x)<sup>1/2</sup> dx, use (as the previous tutor suggested) the fact that cos 2x = 2 cos<sup>2</sup> - 1:
(cos 2x) / (1 + cos 2x)<sup>1/2</sup>
. . .= ( 2 cos<sup>2</sup> x - 1 ) / ( 2 cos<sup>2</sup> x )<sup>1/2</sup>
. . .= ( 2 cos<sup>2</sup> x - 1 ) / 2<sup>1/2</sup> cos x
. . .= ( 2 cos<sup>2</sup> x ) / ( 2<sup>1/2</sup> cos x )- 1 / ( 2<sup>1/2</sup>cos x )
. . .= 2<sup>1/2</sup> cos x - 2<sup>1/2</sup> sec x
integral ( 2<sup>1/2</sup> cos x - 2<sup>1/2</sup> sec x ) dx
. . .= 2<sup>1/2</sup> sin x - 2<sup>1/2</sup> ln | sec x + tan x | + C
If you have questions, please reply. Thank you! :wink:
Re: Integration Problem
Hello, Lizzy!
Identity:
. . \(\displaystyle \cos^2x\:=\:\frac{1\.+\,\cos2x}{2}\;\;\Rightarrow\;\;\cos2x \:=\:2\cdot\cos^2x\,-\,1 \;\;\Rightarrow\;\;1\,+\,\cos 2x\:=\:2\cdot\cos^2x\)
The integral becomes: \(\displaystyle \L\:\int\frac{2\cdot\cos^2x\,-\,1}{\sqrt{2\cdot\cos^2x}}\,dx\:=\:\frac{1}{\sqrt{2}}\int\frac{2\cdot\cos^2x\,-\,1}{\cos x}\,dx\)
And we have: \(\displaystyle \:\frac{\sqrt{2}}{2}\L\int\)\(\displaystyle (2\cdot\cos x\,-\,\sec x)\,dx\;=\;\frac{\sqrt{2}}{2}\left(2\cdot\sin x\,-\,\ln|\sec x\,+\,\tan x|\right)\,+\,C\)
Edit: I just noticed that this is Dora's solution . . .
Hello, Lizzy!
\(\displaystyle \L\int \frac{\cos2x}{\sqrt{1\,+\,\cos2x}}\,dx\)
Identity:
. . \(\displaystyle \cos^2x\:=\:\frac{1\.+\,\cos2x}{2}\;\;\Rightarrow\;\;\cos2x \:=\:2\cdot\cos^2x\,-\,1 \;\;\Rightarrow\;\;1\,+\,\cos 2x\:=\:2\cdot\cos^2x\)
The integral becomes: \(\displaystyle \L\:\int\frac{2\cdot\cos^2x\,-\,1}{\sqrt{2\cdot\cos^2x}}\,dx\:=\:\frac{1}{\sqrt{2}}\int\frac{2\cdot\cos^2x\,-\,1}{\cos x}\,dx\)
And we have: \(\displaystyle \:\frac{\sqrt{2}}{2}\L\int\)\(\displaystyle (2\cdot\cos x\,-\,\sec x)\,dx\;=\;\frac{\sqrt{2}}{2}\left(2\cdot\sin x\,-\,\ln|\sec x\,+\,\tan x|\right)\,+\,C\)
Edit: I just noticed that this is Dora's solution . . .