Definite Integration
Trig Rational
Grade 12
Question:
<p>Evaluate \(\displaystyle\int_0^{\pi/2}\frac{dx}{2+\cos x}\) [JEE Main 2018]</p>
<li>\(\dfrac{\pi}{3\sqrt{3}}\)</li>
<li>\(\dfrac{\pi}{3}\)</li>
<li>\(\dfrac{\pi}{6}\)</li>
<li>\(\dfrac{\pi}{2\sqrt{3}}\)</li>
Step-by-Step Solution
Key Concept: Weierstrass t=tan(x/2): cos x=(1-t^2)/(1+t^2), dx=2dt/(1+t^2). Integral becomes \int_0^1 2dt/(2(1+t^2)+(1-t^2)) = \int_0^1 2dt/(t^2+3).
<div class='solution'>
<p>Let $t=\tan(x/2)$: $\cos x=\frac{1-t^2}{1+t^2}$, $dx=\frac{2dt}{1+t^2}$. Limits: $x=0\to t=0$; $x=\pi/2\to t=1$.</p>
<p>$$I=\int_0^1\frac{2dt/(1+t^2)}{2+\frac{1-t^2}{1+t^2}}=\int_0^1\frac{2dt}{2(1+t^2)+1-t^2}=\int_0^1\frac{2dt}{t^2+3}$$</p>
<p>$$=\frac{2}{\sqrt{3}}\left[\arctan\frac{t}{\sqrt{3}}\right]_0^1=\frac{2}{\sqrt{3}}\cdot\arctan\frac{1}{\sqrt{3}}=\frac{2}{\sqrt{3}}\cdot\frac{\pi}{6}=\boxed{\frac{\pi}{3\sqrt{3}}}$$</p>
</div>
Correct Answer: A