Q.Find $\int \dfrac{\sin 2x \cos 2x\, dx}{\sqrt{9 - \cos^4(2x)}}$

Concept: U Substitution — the denominator’s form $\sqrt{9 - (\cos^2 2x)^2}$ suggests setting $u = \cos^2 2x$.

Step 1: Let $u = \cos^2 2x$. Then $du = 2\cos 2x (-\sin 2x \cdot 2)\, dx = -4 \sin 2x \cos 2x \, dx$.

Hence $\sin 2x \cos 2x \, dx = -\frac{du}{4}$.

Step 2: The integral becomes

$\displaystyle \int \frac{-\frac{du}{4}}{\sqrt{9 - u^2}} = -\frac{1}{4} \int \frac{du}{\sqrt{9 - u^2}}$. …

Let $u=\cos^2(2x)$. Then

so $\sin(2x)\cos(2x)\,dx=-\tfrac14\,du$. Also $\cos^4(2x)=u^2$, so $\sqrt{9-\cos^4(2x)}=\sqrt{9-u^2}$. …

Method: Substitute to reach the standard $\int\dfrac{du}{\sqrt{a^2-u^2}}$ form

Use this when a root $\sqrt{a^2-[\,h(x)\,]^2}$ sits in the denominator and a constant multiple of $h(x)$'s derivative multiplies the top — the integral is really a disguised $\arcsin$.

Steps

Step 1: Recognise the $a^2-u^2$ pattern under the root.

Rewrite $9-\cos^4(2x)$ as $3^2-\big(\cos^2 2x\big)^2$, so $a=3$ and the natural inner variable is $u=\cos^2 2x$.

Step 2: Substitute $u$ and compute $du$.

so $\sin 2x\cos 2x\,dx=-\tfrac{du}{4}$ — exactly the numerator (up to the constant $-\tfrac14$). …

Common Mistakes

Mistake 1: Mis-differentiating $u=\cos^2 2x$.

Why it's wrong: there are two chain-rule layers — the square and the $2x$ — giving $du=-4\sin 2x\cos 2x\,dx$; students often get $-2\sin 2x\cos 2x$ or miss the inner $2$. Correct approach: differentiate the outer square, then $\cos 2x$, then $2x$, multiplying all factors.

Mistake 2: Forgetting the negative sign / the $\tfrac14$.

Why it's wrong: $\sin 2x\cos 2x\,dx=-\tfrac{du}{4}$, so the answer carries $-\tfrac14$; dropping the sign or the $\tfrac14$ gives a wrong coefficient. Correct approach: solve $du$ for the exact numerator factor and keep the $-\tfrac14$. …