ferhaouim3
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- Sep 20, 2005
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hi there,
can you find the antiderevative of: square root of (1+cosx), thank you :lol:
can you find the antiderevative of: square root of (1+cosx), thank you :lol:
Antiderevative
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opticaltempest
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- Nov 19, 2005
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We want to find the derivative of \(\displaystyle \sqrt {1 + \cos \left( x \right)} \\)
It helps to rewrite the our original expression as, \(\displaystyle \left( {1 + \cos (x)} \right)^{\frac{1}{2}}
\\)
Do you see how you can apply the general power rule?
EDIT: It was late... I read it as derivative... Sorry
[/b]
It helps to rewrite the our original expression as, \(\displaystyle \left( {1 + \cos (x)} \right)^{\frac{1}{2}}
\\)
Do you see how you can apply the general power rule?
EDIT: It was late... I read it as derivative... Sorry
[/b]Hello, ferhaouim3!
Pay no attention to the man behind the curtain . . .
We have: . \(\displaystyle \L\int \sqrt{1\,+\,\cos(x)}\,dx\)
Recall the identity: .\(\displaystyle cos^2\left(\frac{x}{2}\right) \:= \:\frac{1\,+\,\cos(x)}{2}\;\;\Rightarrow\;\;1\,+\,\cos(x)\:=\:2\cdot\cos^2\left(\frac{x}{2}\right)\)
. . Hence: .\(\displaystyle \sqrt{1\,+\,\cos(x)} \:=\:\sqrt{2}\cdot\cos\left(\frac{x}{2}\right)\)
The integral becomes: .\(\displaystyle \L\sqrt{2}\int\cos\left(\frac{x}{2}\right)\,dx\) . . . got it?
Pay no attention to the man behind the curtain . . .
We have: . \(\displaystyle \L\int \sqrt{1\,+\,\cos(x)}\,dx\)
Recall the identity: .\(\displaystyle cos^2\left(\frac{x}{2}\right) \:= \:\frac{1\,+\,\cos(x)}{2}\;\;\Rightarrow\;\;1\,+\,\cos(x)\:=\:2\cdot\cos^2\left(\frac{x}{2}\right)\)
. . Hence: .\(\displaystyle \sqrt{1\,+\,\cos(x)} \:=\:\sqrt{2}\cdot\cos\left(\frac{x}{2}\right)\)
The integral becomes: .\(\displaystyle \L\sqrt{2}\int\cos\left(\frac{x}{2}\right)\,dx\) . . . got it?