Evaluating $\int_{-1}^{1}\sin^7\left({x}\right) \,dx$ Using $u$-Substitution

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SUMMARY

The integral $\int_{-1}^{1}\sin^7\left({x}\right) \,dx$ evaluates to 0 due to the properties of odd functions. By applying the substitution $u=\cos\left({x}\right)$ and $du=-\sin\left({x}\right)dx$, the integral transforms into $-\int_{-1}^{1} \left(1-{u}^{2}\right)^3\,du$. Since the integrand is an odd function and the limits of integration are symmetric about the origin, the areas under the curve cancel each other, confirming that the integral equals zero.

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karush
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$\int_{-1}^{1}\sin^7\left({x}\right) \,dx$
$u=\cos\left({x}\right)$
$du=-\sin\left({x}\right)$
So
$-\int_{-1}^{1} \left(1-{u}^{2}\right)^3\,du$
So continue?
 
Last edited:
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Your integrand is odd, and the limits symmetric about the origin, so can you think of a shortcut you can use?
 
I presume you mean the areas will cancel out and the answer will be zero for interval given?
 
karush said:
I presume you mean the areas will cancel out and the answer will be zero for interval given?

Yes, the odd-function rule can be directly applied here to state that the given definite integral has a value of 0. :D
 

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