Question Statement

Q17. Evaluate the integral: $\int x^2\sin xdx$

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Background and Explanation

Integration by parts is the reverse application of the product rule for differentiation, used when integrating products of functions—particularly when one function is algebraic (like a polynomial) and the other is trigonometric. When the algebraic function has a power greater than 1, we must apply integration by parts repeatedly, reducing the power each time until the remaining integral becomes straightforward.

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Solution

Let $I=\int x^2\sin xdx$.

Step 1: First application of integration by parts

Following the LIATE rule (Logarithmic, Inverse, Algebraic, Trigonometric, Exponential), we choose the algebraic function as $u$ to differentiate it (which reduces its power):

• Let $u=x^2$, so $\frac{du}{dx}=2x$ or $du=2xdx$
• Let $dv=\sin xdx$, so $v=\int \sin xdx=-\cos x$

Applying the integration by parts formula $\int udv=uv-\int vdu$:

$\begin{array}{rl}I& =x^2(-\cos x)-\int (-\cos x)(2x)dx\\ & =-x^2\cos x+2\int x\cos xdx\end{array}$

Step 2: Second application of integration by parts

The remaining integral $\int x\cos xdx$ still requires integration by parts. Again, we choose the algebraic part as $u$:

• Let $u=x$, so $du=dx$
• Let $dv=\cos xdx$, so $v=\int \cos xdx=\sin x$

Applying the formula again:

$\begin{array}{rl}\int x\cos xdx& =x\sin x-\int \sin xdx\\ & =x\sin x-(-\cos x)\\ & =x\sin x+\cos x\end{array}$

Step 3: Substitute back and simplify

Substituting this result back into our expression for $I$:

$\begin{array}{rl}I& =-x^2\cos x+2(x\sin x+\cos x)+C\\ & =-x^2\cos x+2x\sin x+2\cos x+C\end{array}$

Where $C$ is the constant of integration.

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Key Formulas or Methods Used

• Integration by parts formula: $\int udv=uv-\int vdu$
• LIATE priority rule: For choosing $u$ (priority order: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential)
• Basic trigonometric integrals:
  • $\int \sin xdx=-\cos x+C$
  • $\int \cos xdx=\sin x+C$
• Power rule for differentiation: $\frac{d}{dx}(x^n)=n{x}^{n-1}$

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Summary of Steps

1. Identify the method: Recognize this as a product of polynomial ($x^2$) and trigonometric ($\sin x$) functions requiring integration by parts.
2. First IBP: Set $u=x^2$ and $dv=\sin xdx$ to reduce the polynomial degree, yielding $-x^2\cos x+2\int x\cos xdx$.
3. Second IBP: Apply integration by parts again to $\int x\cos xdx$ with $u=x$ and $dv=\cos xdx$.
4. Evaluate the simpler integral: Compute $\int \sin xdx=-\cos x$ from the second application.
5. Combine results: Substitute back to get $-x^2\cos x+2(x\sin x+\cos x)+C$.
6. Final simplification: Expand and arrange terms to obtain $-x^2\cos x+2x\sin x+2\cos x+C$.