Trigonometric substitution: int cos^5 (17x) dx, -2pi to pi/2
- Thread starter flakine
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Here's a nice little general formula for the integrals of \(\displaystyle cos^{5}(nx)\)
\(\displaystyle \L\\\int{cos^{5}(nx)}dx=\frac{sin(nx)(3cos^{4}(nx)+4cos^{2}(nx)+8)}{15n}\)
You have n=17
If you must integrate it all the long arduous way, you could start with the identity
\(\displaystyle 1-sin^{2}(17x)=cos^{2}(17x)\)
Then you have:
\(\displaystyle \L\\\int_{-2\pi}^{\frac{\pi}{2}}(1-sin^{2}(17x))^{2}cos(17x)dx\)
\(\displaystyle \L\\\int{cos^{5}(nx)}dx=\frac{sin(nx)(3cos^{4}(nx)+4cos^{2}(nx)+8)}{15n}\)
You have n=17
If you must integrate it all the long arduous way, you could start with the identity
\(\displaystyle 1-sin^{2}(17x)=cos^{2}(17x)\)
Then you have:
\(\displaystyle \L\\\int_{-2\pi}^{\frac{\pi}{2}}(1-sin^{2}(17x))^{2}cos(17x)dx\)
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- Apr 12, 2005
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I'm rather fond of some symmetry exploitation.
I assumed you meant -\(\displaystyle \pi\)/2, not -2\(\displaystyle \pi\)
\(\displaystyle \L\,2*\int_{0}^{\frac{\pi}{34}}{cos^{5}(17x)}\,dx\)
Unless you REALLY meant -2\(\displaystyle \pi\), then it's just
\(\displaystyle \L\,\int_{0}^{\frac{\pi}{34}}{cos^{5}(17x)}\,dx\)
Of course, that doesn't simplify the antiderivative any. It certainly does simplify life in the numerical world.
I assumed you meant -\(\displaystyle \pi\)/2, not -2\(\displaystyle \pi\)
\(\displaystyle \L\,2*\int_{0}^{\frac{\pi}{34}}{cos^{5}(17x)}\,dx\)
Unless you REALLY meant -2\(\displaystyle \pi\), then it's just
\(\displaystyle \L\,\int_{0}^{\frac{\pi}{34}}{cos^{5}(17x)}\,dx\)
Of course, that doesn't simplify the antiderivative any. It certainly does simplify life in the numerical world.