MHB Integral $$\sin^7\left({x}\right)$$ from -1 to 1: Solution Explained

[karush]

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

From the graph of this it's apparent the answer is 0

But step wise. I did this

$$\int_{-1}^{1}(\sin^2\left({x}\right) )^3\sin\left({x}\right)\,dx$$

With
$$\d{}{x}\left(\cos^2\left(x\right)-1\right)=2\sin\left({x}\right)\cos\left({x}\right)$$

Couldn't get the derivative needed?

 

[MarkFL]

The integrand is an odd-function ($f(-x)=-f(x)$), and the limits are symmetric about $x=0$, and so yes, the immediate result (by the odd-function rule) is that the definite integral evaluates to zero. :)

 

[karush]

So this this can be determined by the interval and the odd power...

 

[MarkFL]

So this this can be determined by the interval and the odd power...

Well, it is the fact that sine is odd ($\sin(-x)=-\sin(x)$), and the exponent is odd, and so the integrand is an odd function overall. If $f$ is an odd function (and continuous on $(-a,a)$), then we must have:

$$\int_{-a}^0 f(x)\,dx=-\int_0^{a} f(x)\,dx$$

or

$$\int_{-a}^0 f(x)\,dx+\int_0^{a} f(x)\,dx=0$$

or

$$\int_{-a}^a f(x)\,dx=0$$

 

[Greg]

$$\int_{-1}^{1}(\sin^2\left({x}\right) )^3\sin\left({x}\right)\,dx$$

$$\int_{-1}^1(1-\cos^2x)^3\sin x\,dx$$

Now make the sub $u=-\cos x$ and see what happens when you recalculate the bounds of integration.

 

[karush]

so basically it was double substitution on this one

 

[Greg]

I don't see a more direct way of doing it.

 

[Guest2]

Let $x = -t$ and see what happens :)