Q.Integrate the following function: $\sin^3 x \cos^3 x$

Write everything with the double angle: $\sin x\cos x=\frac12\sin 2x$.

For $\int\sin^3 2x\,dx$, write $\sin^3 2x=(1-\cos^2 2x)\sin 2x$ and let $u=\cos 2x$, $du=-2\sin 2x\,dx$:

Multiply by $\frac18$: …

Compress with the double angle

Both factors share the same power, so group them: $\sin^3 x\cos^3 x=(\sin x\cos x)^3$. Using $\sin x\cos x=\frac12\sin 2x$,

so $\int\sin^3 x\cos^3 x\,dx=\frac18\int\sin^3 2x\,dx$.

Odd power of sine: save one factor

$\sin^3 2x=\sin^2 2x\cdot\sin 2x=(1-\cos^2 2x)\sin 2x$. The spare $\sin 2x$ is perfect for the substitution $u=\cos 2x$, since $du=-2\sin 2x\,dx$, i.e. $\sin 2x\,dx=-\frac12\,du$:

Restore the $\frac18$

$\int\sin^3 x\cos^3 x\,dx=\frac18\left(-\frac12\cos 2x+\frac16\cos^3 2x\right)+C=-\frac{1}{16}\cos 2x+\frac{1}{48}\cos^3 2x+C.$ …

Method: Equal odd powers of $\sin$ and $\cos$ — compress with the double angle

For $\sin^m x\cos^n x$ where both powers are equal and odd, group them as $(\sin x\cos x)^m$, use $\sin x\cos x=\tfrac12\sin 2x$, then handle the resulting odd power of $\sin 2x$ by the save-one-factor substitution.

Steps

Step 1: Group the equal powers.

Step 2: Collapse with the double-angle identity.

Step 3: Integrate the odd power of $\sin 2x$ by substitution. …

Common Mistakes

Mistake 1: Trying the power rule on $\sin^3 x\cos^3 x$ as if it were a simple power.

Why it's wrong: neither $\sin x$ nor $\cos x$ has its derivative sitting alone as a factor of the whole product, so no single-step substitution or power rule applies. Correct approach: compress via $(\sin x\cos x)^3=\tfrac18\sin^3 2x$, then substitute $u=\cos 2x$.

Mistake 2: Forgetting the $-2$ in $du=-2\sin 2x\,dx$ when substituting. …