Q.Integrate the following function: $\sqrt{\sin 2x} \cos 2x$

Concept: U Substitution — we let $u$ equal the expression inside the square root so that its derivative appears as a factor.

Let $u = \sin 2x$. Then $du = 2\cos 2x \, dx$, so $\cos 2x \, dx = \frac{du}{2}$.

The integral becomes: …

Why U-Substitution Works Here

When you see a function like $\sqrt{\sin 2x} \cos 2x$, your first instinct should be to look for a function and its derivative hiding inside. The derivative of $\sin 2x$ is $2\cos 2x$ — and we have $\cos 2x$ sitting right there, just missing a factor of 2. That’s the classic signal for substitution: the integrand is a product of a composite function and the derivative of its inner part (up to a constant).

The square root $\sqrt{\sin 2x}$ is really $(\sin 2x)^{1/2}$, so we’re integrating something of the form $(\text{inside})^{1/2} \times (\text{derivative of inside})$. That’s a power rule in disguise.

Step-by-Step Solution

1. Set up the substitution.

Let $u = \sin 2x$. Then differentiate:

2. Solve for the piece we have.

Our integrand has $\cos 2x \, dx$, not $2\cos 2x \, dx$. So divide both sides by 2:

3. Rewrite the integral in terms of $u$.

The original integral is

Substituting $u$ and $\frac{du}{2}$ gives:

4. Integrate using the power rule. …

Method: Substitute the inside of the root when its derivative is present

Use this when a root of a trig function multiplies that function's derivative, e.g. $\sqrt{\sin 2x}\,\cos 2x$.

Steps

Step 1: Check the derivative pairing.

$\dfrac{d}{dx}(\sin 2x)=2\cos 2x$, and $\cos 2x$ is present in the integrand.

Step 2: Substitute $u=\sin 2x$. …

Common Mistakes

Mistake 1: Forgetting the $\dfrac12$ from $du=2\cos 2x\,dx$.

Why it's wrong: with $u=\sin 2x$, $\cos 2x\,dx=\dfrac{du}{2}$; the factor $2$ comes from differentiating $\sin 2x$. Correct approach: account for the chain-rule factor before integrating.

Mistake 2: Choosing the wrong $u$.

Why it's wrong: the substitution must be $u=\sin 2x$ (whose derivative $\cos 2x$ is present), not $u=\cos 2x$. Correct approach: pick $u$ so $du$ appears in the integrand. …