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

The key idea is integration by parts, which reverses the product rule. We choose $u$ and $dv$ so that $du$ is simpler and $v$ is easy to find.

Let $u = x$ and $dv = \sin x \, dx$. Then $du = dx$ and $v = -\cos x$.

Applying integration by parts:

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

Why integration by parts?

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

Rearranging and integrating both sides gives the formula:

The art is in choosing which part to call $u$ and which to call $dv$. A good rule of thumb: pick $u$ to be the function that simplifies when differentiated, and $dv$ to be the one that doesn't get more complicated when integrated.

Here, $x$ becomes simpler when differentiated (it becomes $1$), while $\sin x$ integrates nicely to $-\cos x$. So we set:

• $u = x$
• $dv = \sin x \, dx$

Step-by-step

1. Differentiate $u$ and integrate $dv$

(We can ignore the constant of integration here; it will be absorbed at the end.)

2. Apply the integration by parts formula

3. Simplify the expression

The minus signs need care:

4. Integrate $\cos x$

5. Write the final result

Method: Integration by parts (polynomial $\times$ trig), choose $u$ by ILATE

Whenever you integrate a product of different kinds of functions — a polynomial and a trig function — reverse the product rule with integration by parts.

Steps

Step 1: Recall the formula and the choice rule.

Pick $u$ by ILATE (Inverse, Logarithmic, Algebraic, Trigonometric): the one earlier in the list becomes $u$. For polynomial $\times$ trig, the algebraic (polynomial) part is $u$, the trig part is $dv$.

Step 2: Why that choice.

Differentiating the polynomial lowers its degree (toward a constant), so the leftover integral $\int v\,du$ is genuinely simpler. Choosing the trig part as $u$ instead raises the polynomial's degree — the integral gets worse.

Step 3: Differentiate $u$, integrate $dv$.

With $u=x$: $du=dx$, and $v=\int\sin x\,dx=-\cos x$.

Step 4: Assemble and finish.

Track the double negative $-\int(-\cos x)\,dx = +\int\cos x\,dx$ carefully.

Common Mistakes

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

Why it's wrong: Then $v=\frac{x^2}{2}$ and the leftover $\int\frac{x^2}{2}\cos x\,dx$ is harder than the original. Correct approach: By ILATE the algebraic $x$ is $u$; the polynomial degree should drop, not rise.

Mistake 2: Sign slip on the double negative.

Why it's wrong: With $v=-\cos x$, the formula gives $-x\cos x-\int(-\cos x)\,dx=-x\cos x+\int\cos x\,dx$; students often write $-\int\cos x\,dx$. Correct approach: Track $uv-\int v\,du$ with $v=-\cos x$ carefully, so the remaining integral is $+\int\cos x\,dx$.

Mistake 3: Stopping at $-x\cos x$ without integrating the leftover.

Why it's wrong: The $\int\cos x\,dx=\sin x$ term is part of the answer. Correct approach: Finish to $-x\cos x+\sin x+C$.