Definite Integration
Trigonometric Powers
Grade None
Question:
<p>Evaluate \(\displaystyle\int_0^{\pi/2}\sin^4 x\,dx\) [JEE Main 2020]</p>
<li>\(\dfrac{3\pi}{16}\)</li>
<li>\(\dfrac{\pi}{4}\)</li>
<li>\(\dfrac{3\pi}{8}\)</li>
<li>\(\dfrac{\pi}{16}\)</li>
Step-by-Step Solution
Key Concept: Wallis: \int_0^(\pi/2) sin^(2n) x dx = [(2n-1)\!\!/(2n)\!\!] \cdot (\pi/2). For n=2: (3 \cdot 1)/(4 \cdot 2) \cdot (\pi/2) = (3/8) \cdot (\pi/2) = 3\pi/16.
<div class='solution'>
<p>Using Wallis: \(\int_0^{\pi/2}\sin^4 x\,dx=\frac{3\!\!}{4\!\!}\cdot\frac{\pi}{2}=\frac{3\cdot1}{4\cdot2}\cdot\frac{\pi}{2}=\frac{3}{8}\cdot\frac{\pi}{2}=\frac{3\pi}{16}\)</p>
<p><em>Alternatively:</em> \(\sin^4 x=\frac{3-4\cos 2x+\cos 4x}{8}\). \(\int_0^{\pi/2}=\frac{1}{8}\left[3x-2\sin 2x+\frac{\sin 4x}{4}\right]_0^{\pi/2}=\frac{1}{8}\cdot\frac{3\pi}{2}=\frac{3\pi}{16}\)</p>
</div>
Correct Answer: A