Definite Integration
Trigonometric + King
Grade 12
Question:
<p>Evaluate \(\displaystyle\int_0^{\pi}x\sin^3 x\,dx\)</p>
<li>\(\dfrac{\pi}{2}\)</li>
<li>\(\dfrac{2\pi}{3}\)</li>
<li>\(\dfrac{2\pi}{3}\)</li>
<li>\(\dfrac{\pi}{3}\)</li>
Step-by-Step Solution
Key Concept: King's rule: I = \int_0^\pi (\pi-x)sin^3x dx, add to get 2I = \pi\int_0^\pi sin^3x dx. Use sin^3x = (3sinx - sin3x)/4.
<div class='solution'>
<p>Let $I=\int_0^\pi x\sin^3 x\,dx$. King ($x\to\pi-x$): $\sin(\pi-x)=\sin x$, so</p>
<p>$$I = \int_0^\pi(\pi-x)\sin^3 x\,dx$$</p>
<p>Add: $2I = \pi\int_0^\pi\sin^3 x\,dx$.</p>
<p>$\sin^3 x = \dfrac{3\sin x-\sin 3x}{4}$</p>
<p>$$\int_0^\pi\sin^3 x\,dx = \frac{1}{4}\left[-3\cos x+\frac{\cos 3x}{3}\right]_0^\pi = \frac{1}{4}\left[(3-\frac{1}{3})-(-3+\frac{1}{3})\right] = \frac{1}{4}\cdot\frac{16}{3} = \frac{4}{3}$$</p>
<p>$$2I = \pi\cdot\frac{4}{3}\Rightarrow I = \boxed{\frac{2\pi}{3}}$$</p>
</div>
Correct Answer: C