Q.If $I_n = \int \sin^n x\,dx$ for an integer $n \geq 2$, then show that $I_n = \frac{-\sin^{n-1}x\cos x}{n} + \frac{(n-1)}{n}I_{n-2}$.

Sine Power Reduction

How do you integrate something like $\sin^2 x$? If you try the ordinary power rule you get stuck — there is no simple antiderivative you can just write down for a squared trig function. Power reduction is the standard fix: rewrite an even power of $\sin x$ as a constant plus a cosine of a larger angle, turning an un-integrable lump into terms you already know how to handle.

The Core Idea

Start from the double-angle identity for cosine:

Solve this for $\sin^2 x$:

Notice what happened: the power dropped from $2$ to $1$. On the right we only have a constant and a single cosine term — and $\int \cos(kx)\,dx = \frac{1}{k}\sin(kx)$ is easy. That is the whole point of "power reduction": trade a squared trig function for the double angle.

Why It Works

The identity is exact, not an approximation — it is just the $\cos 2x = 1 - 2\sin^2 x$ relation rearranged. So $\sin^2 x$ and $\frac{1-\cos 2x}{2}$ are literally the same function; replacing one with the other never changes the value, only the form, into a form that integrates cleanly.

Using It

Higher even powers are handled by applying the trick again. For example:

The leftover $\cos^2 2x$ is still a square, so reduce it once more with $\cos^2 2x = \frac{1 + \cos 4x}{2}$. Each pass lowers the power until everything is linear in cosine.