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

The key idea is Integration by Parts, which reverses the product rule. We choose $u$ and $dv$ so that the new integral is simpler.

Let $u = x$ and $dv = \sin 3x \, dx$. Then $du = dx$ and $v = \int \sin 3x \, dx = -\frac{1}{3} \cos 3x$.

Apply the formula $\int u \, dv = uv - \int v \, du$:

Simplify and integrate the remaining term:

Why integration by parts?

When you see a product of two different kinds of functions — here, a polynomial ($x$) and a trigonometric function ($\sin 3x$) — the standard tool is integration by parts. It comes from the product rule for derivatives:

Rearranging and integrating gives:

The trick is to pick $u$ and $dv$ so that the new integral $\int v \, du$ is simpler than the original. For $x \sin 3x$, we want $u$ to be something that simplifies when differentiated (like $x$, which becomes $1$), and $dv$ to be something we can integrate easily (like $\sin 3x$).

Step-by-step solution

1. Set up the parts.

Let $u = x$ and $dv = \sin 3x \, dx$.

2. Differentiate $u$ and integrate $dv$.

• $du = dx$
• $v = \int \sin 3x \, dx = -\frac{1}{3} \cos 3x$

3. Apply the integration by parts formula.

Substitute:

4. Simplify the expression.

5. Integrate $\cos 3x$.

So:

6. Write the final result.

Method: Integration by parts with a scaled angle $\sin(ax)$

Same by-parts idea as $x\sin x$, but now the trig function has a coefficient $a$ inside, which introduces factors of $\dfrac1a$ at each integration.

Steps

Step 1: Choose $u$ and $dv$ by ILATE.

Algebraic before trigonometric: $u=x$ (differentiates to $1$), $dv=\sin(ax)\,dx$.

Step 2: Integrate $dv$ using the chain-rule factor.

Each integration of a scaled-angle function multiplies by $\dfrac1a$; forgetting this is the usual error.

Step 3: Apply $\int u\,dv=uv-\int v\,du$.

Step 4: Finish and collect the $\dfrac1a$ factors.

For $a=3$ this gives the $\dfrac19\sin 3x$ term — note it is $\dfrac{1}{a^2}$, not $\dfrac1a$.

Common Mistakes

Mistake 1: Ignoring the $\tfrac1a$ factor for $\sin(3x)$.

Why it's wrong: $\int\sin 3x\,dx=-\tfrac13\cos 3x$ and $\int\cos 3x\,dx=\tfrac13\sin 3x$; treating them as if $a=1$ gives wrong coefficients. Correct approach: Divide by the inner coefficient $a=3$ at each integration step.

Mistake 2: Writing $\tfrac13\sin 3x$ instead of $\tfrac19\sin 3x$.

Why it's wrong: The final $\sin 3x$ term picks up $\tfrac1a$ twice (once from $v$, once from integrating $\cos 3x$), giving $\tfrac{1}{a^2}=\tfrac19$. Correct approach: Expect $\tfrac{1}{a^2}$ on the $\sin$ term for $x\sin(ax)$.

Mistake 3: Swapping $u$ and $dv$.

Why it's wrong: Taking $u=\sin 3x$ makes the next integral worse. Correct approach: Let the polynomial $x$ be $u$ (ILATE).