How to Integrate ∫sin^3(x)cos^4(x)dx

In this video, I demonstrate how to integrate sin³xcos⁴x by reserving a factor of sin(x) and converting the integrand into powers of cos(x). This then turns this integral into a simple power integral.

The first step is to write sin³(x) as sin(x)sin²(x).

Then by the Pythagorean identity, sin²(x) = 1 - cos²(x). So the integral becomes:

\int\sin x\left [ 1-\cos^2x \right ]\cos^4x\mathrm{d}x

Rearranging a little bit, we get:

\int\left [ 1-\cos^2x \right ]\cos^4x\sin x\mathrm{d}x

Now, if we make the substitution:

u = \cos x

The derivative is du/dx = -sin(x). And thus:

-\mathrm{d}u=\sin x\mathrm{d}x

Substituting, we get

\int \left [ 1 - u^2 \right ]u^4\cdot -\mathrm{d}u = \int \left [ u^6-u^4 \right ]\mathrm{d}u

This is then a simple power integral.

This approach can be applied to any integral ∫sin^m(x)cosⁿ(x)dx where the powers m and n are odd and even respectively.

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