Q.Find $\int x \cos x\, dx$

The key idea is integration by parts, which reverses the product rule. We choose $u = x$ (so $du = dx$) and $dv = \cos x\, dx$ (so $v = \sin x$).

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

The remaining integral is standard: $\int \sin x\, dx = -\cos x + C$.

Thus:

Why integration by parts?

When you see a product of two different kinds of functions — here $x$ (algebraic) and $\cos x$ (trigonometric) — there’s no simple reverse derivative. The product rule for differentiation says $(uv)' = u'v + uv'$, so rearranging gives:

This is integration by parts. The trick is to pick $u$ so that $du$ is simpler, and $dv$ so that $v$ is easy to integrate.

Step-by-step solution

1. Choose $u$ and $dv$

Let $u = x$ and $dv = \cos x \, dx$.

Why? Because $du = dx$ becomes simpler (the power drops), and $v = \sin x$ is easy to integrate.

2. Compute $du$ and $v$

3. Apply the integration by parts formula

4. Integrate the remaining term

So:

5. Check by differentiating

Differentiate $x \sin x + \cos x$:

• Derivative of $x \sin x$: using product rule, $1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x$
• Derivative of $\cos x$: $-\sin x$

Sum: $\sin x + x \cos x - \sin x = x \cos x$ — matches the integrand. Perfect.

Method: Integration by Parts (Algebraic × Trigonometric)

Use this when the integrand is a polynomial times a trig function, where differentiating the polynomial simplifies it.

Steps

Step 1: Apply the by-parts formula with the right choice.

By LIATE, choose $u$ as the algebraic factor (so $du$ is simpler) and $dv$ as the trig factor. For $\int x\cos x\,dx$, take $u=x$, $dv=\cos x\,dx$.

Step 2: Compute $du$ and $v$.

Here $du=dx$ and $v=\sin x$. The formula gives $x\sin x-\int\sin x\,dx$.

Step 3: Finish the simpler integral.

$\int\sin x\,dx=-\cos x$, so the result is $x\sin x+\cos x+C$. Verify by differentiating.

Common Mistakes

Mistake 1: Choosing $u=\cos x$, $dv=x\,dx$.

Why it's wrong: then $v=\frac{x^2}{2}$ makes the new integral harder, not easier. Correct approach: pick $u=x$ (algebraic) so its derivative is a constant.

Mistake 2: Sign error integrating $\int\sin x\,dx$.

Why it's wrong: $\int\sin x\,dx=-\cos x$, and combined with the formula's minus sign the term becomes $+\cos x$. Correct approach: track both minus signs.

Mistake 3: Forgetting the $-\int v\,du$ term entirely.

Why it's wrong: writing only $uv=x\sin x$ omits half the answer. Correct approach: always subtract $\int v\,du$.