Q.Find $\int \cos 6x \sqrt{1 + \sin 6x}\, dx$

Concept: U Substitution — we set $u$ equal to the expression inside the square root because its derivative is a constant multiple of $\cos 6x$.

Let $u = 1 + \sin 6x$. Then

$\frac{du}{dx} = 6\cos 6x$, so $\cos 6x\, dx = \frac{du}{6}$.

The integral becomes

$\int \sqrt{u} \cdot \frac{du}{6} = \frac{1}{6} \int u^{1/2}\, du$.

Integrate:

$\frac{1}{6} \cdot \frac{u^{3/2}}{3/2} = \frac{1}{6} \cdot \frac{2}{3} u^{3/2} = \frac{1}{9} u^{3/2}$.

Substitute back $u = 1 + \sin 6x$ and add the constant of integration.

The key to this problem is noticing that the derivative of $\sin 6x$ is $6\cos 6x$, which is almost exactly the other factor in the integrand. That’s a perfect setup for u-substitution — we let the expression inside the square root be $u$, so that its derivative handles the $\cos 6x$ term.

Let’s walk through it.

1. Choose the substitution. The square root $\sqrt{1 + \sin 6x}$ is the complicated part. Set

Then differentiate:

2. Rewrite the integral in terms of $u$. The original integral is $\int \cos 6x \sqrt{1 + \sin 6x}\, dx$. We have $\sqrt{1 + \sin 6x} = \sqrt{u}$, and $\cos 6x\, dx = \frac{du}{6}$. So the integral becomes:

3. Integrate with respect to $u$. Using the power rule:

4. Substitute back. Replace $u$ with $1 + \sin 6x$:

Method: $u$-substitution — spotting a function beside its own derivative

Use this whenever the integrand is (a function of an inner expression) $\times$ (a constant multiple of that inner expression's derivative) — the classic reverse-chain-rule setup.

Steps

Step 1: Identify the inner function $u$.

Look for the "messy" inner piece — here the quantity under the root. Set $u$ equal to it. A good sign you have picked the right $u$: differentiating it reproduces the other factor in the integrand up to a constant.

Step 2: Differentiate to get $du$, and solve for the factor you need.

Because differentiating a bracketed argument brings out a chain-rule constant, express the leftover factor times $dx$ as a multiple of $du$ (e.g. $\cos 6x\,dx=\tfrac{du}{6}$).

Step 3: Rewrite the whole integral in $u$ and integrate.

Every $x$ and $dx$ must disappear. You are left with a simple power (or standard form) in $u$, integrated by the power rule $\int u^{n}\,du=\dfrac{u^{n+1}}{n+1}+C$.

Step 4: Substitute back and (optionally) verify.

Replace $u$ by $g(x)$ and add $+C$. Differentiate your answer mentally to confirm it reproduces the original integrand — the chain-rule constant must come back out correctly.

Common Mistakes

Mistake 1: Forgetting the chain-rule constant when forming $du$.

Why it's wrong: $\dfrac{d}{dx}\sin 6x=6\cos 6x$, not $\cos 6x$, so $du=6\cos 6x\,dx$ and $\cos 6x\,dx=\tfrac{du}{6}$. Dropping the $6$ makes the final coefficient wrong (you'd get $\tfrac16$ instead of $\tfrac19$). Correct approach: always differentiate the full inner function including its inner coefficient.

Mistake 2: Mishandling the power rule on $u^{1/2}$.

Why it's wrong: $\int u^{1/2}\,du=\dfrac{u^{3/2}}{3/2}=\tfrac23 u^{3/2}$; students often write $\tfrac12 u^{3/2}$ or $u^{3/2}$. Correct approach: add $1$ to the exponent and divide by the new exponent $\tfrac32$.

Mistake 3: Omitting the constant of integration.

Why it's wrong: this is an indefinite integral, so $+C$ is part of the answer. Correct approach: always append $+C$ after back-substituting.