needing to integrate (cos(a sin(a)))^6
- Thread starter Knight
- Start date
Re: Help
Hello, Knight!
You don't seem to have a variable.
I'll take a guess at what you meant . . .
Let \(\displaystyle \theta \:=\:\arcsin x\;\;\Rightarrow\;\;\sin\theta \:=\:x\)
Then \(\displaystyle \theta\) is an angle in a right triangle
. . with: \(\displaystyle opp\,=\,x,\;hyp\,=\,1\)
Hence: \(\displaystyle \,adj\,=\,\sqrt{1\,-\,x^2}\,\) and \(\displaystyle \,\cos\theta \,=\,\sqrt{1\,-\,x^2}\)
So we have: \(\displaystyle \L\:\int\)\(\displaystyle cos^6\theta\,d\theta \;=\;\L\int\)\(\displaystyle \left(\sqrt{1\,-\,x^2}\right)^6\,dx \;=\;\L\int\)\(\displaystyle (1\,-\,x^2)^3\,dx\)
. . \(\displaystyle = \;\L\int\)\(\displaystyle (1\,-\,3x^2\,+\,3x^4\,-\,x^6)\,dx\)
And you can finish it now . . .
Hello, Knight!
You don't seem to have a variable.
I'll take a guess at what you meant . . .
Let \(\displaystyle \theta \:=\:\arcsin x\;\;\Rightarrow\;\;\sin\theta \:=\:x\)
Then \(\displaystyle \theta\) is an angle in a right triangle
. . with: \(\displaystyle opp\,=\,x,\;hyp\,=\,1\)
Hence: \(\displaystyle \,adj\,=\,\sqrt{1\,-\,x^2}\,\) and \(\displaystyle \,\cos\theta \,=\,\sqrt{1\,-\,x^2}\)
So we have: \(\displaystyle \L\:\int\)\(\displaystyle cos^6\theta\,d\theta \;=\;\L\int\)\(\displaystyle \left(\sqrt{1\,-\,x^2}\right)^6\,dx \;=\;\L\int\)\(\displaystyle (1\,-\,x^2)^3\,dx\)
. . \(\displaystyle = \;\L\int\)\(\displaystyle (1\,-\,3x^2\,+\,3x^4\,-\,x^6)\,dx\)
And you can finish it now . . .
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