The value of the integral \int\limits_ - \pi \over 4^\pi \over 4 x + \pi \over 4 \over 2 - \cos 2xdx is :

Definite Integration PYQ — JEE Main 2023 · 1 Feb · Shift 2 · Q30 with Solution

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Correct answer: A

Let $I=\int_{-\pi/4}^{\pi/4} \frac{x+\pi/4}{2-\cos 2x}\,dx$ We need to evaluate this definite integral.
Split the integral: $I=\int_{-\pi/4}^{\pi/4} \frac{x}{2-\cos 2x}\,dx+\frac{\pi}{4}\int_{-\pi/4}^{\pi/4} \frac{dx}{2-\cos 2x}$ So, $I=I_1+I_2$ where

\qquad I_2=\frac{\pi}{4}\int_{-\pi/4}^{\pi/4} \frac{dx}{2-\cos 2x}$$ 3. Evaluate $I_1$ using symmetry. Since $\cos 2x$ is an even function, $2-\cos 2x$ is even. Hence $$\frac{x}{2-\cos 2x}$$ is an odd function (odd/even = odd). Therefore, over symmetric limits $[-a,a]$, $$I_1=\int_{-\pi/4}^{\pi/4} \frac{x}{2-\cos 2x}\,dx=0$$ Thus, $$I=I_2=\frac{\pi}{4}\int_{-\pi/4}^{\pi/4} \frac{dx}{2-\cos 2x}$$ 4. Now compute $$J=\int_{-\pi/4}^{\pi/4} \frac{dx}{2-\cos 2x}$$ Use the substitution $$t=\tan x, \qquad dx=\frac{dt}{1+t^2}, \qquad \cos 2x=\frac{1-t^2}{1+t^2}$$ Also, when $x=-\pi/4$, $t=-1$; and when $x=\pi/4$, $t=1$. Then $$2-\cos 2x=2-\frac{1-t^2}{1+t^2} =\frac{2(1+t^2)-(1-t^2)}{1+t^2} =\frac{1+3t^2}{1+t^2}$$ Therefore, $$\frac{dx}{2-\cos 2x} =\frac{\frac{dt}{1+t^2}}{\frac{1+3t^2}{1+t^2}} =\frac{dt}{1+3t^2}$$ So, $$J=\int_{-1}^{1} \frac{dt}{1+3t^2}$$ 5. Evaluate $J$: $$J=\int_{-1}^{1} \frac{dt}{1+3t^2} =\frac{1}{\sqrt{3}}\left[\tan^{-1}(\sqrt{3}t)\right]_{-1}^{1}$$ $$J=\frac{1}{\sqrt{3}}\left(\tan^{-1}(\sqrt{3})-\tan^{-1}(-\sqrt{3})\right)$$ $$J=\frac{1}{\sqrt{3}}\left(\frac{\pi}{3}+\frac{\pi}{3}\right) =\frac{2\pi}{3\sqrt{3}}$$ 6. Hence, $$I=\frac{\pi}{4}\cdot \frac{2\pi}{3\sqrt{3}} =\frac{\pi^2}{6\sqrt{3}}$$ 7. Compare with the options: - A: $\dfrac{\pi^2}{6\sqrt{3}}$ ✅ - B: $\dfrac{\pi^2}{6}$ - C: $\dfrac{\pi^2}{3\sqrt{3}}$ - D: $\dfrac{\pi^2}{12\sqrt{3}}$ Therefore, the correct option is **A**.

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Correct answer: A

Let $I=\int_{-\pi/4}^{\pi/4} \frac{x+\pi/4}{2-\cos 2x}\,dx$ We need to evaluate this definite integral.
Split the integral: $I=\int_{-\pi/4}^{\pi/4} \frac{x}{2-\cos 2x}\,dx+\frac{\pi}{4}\int_{-\pi/4}^{\pi/4} \frac{dx}{2-\cos 2x}$ So, $I=I_1+I_2$ where

\qquad I_2=\frac{\pi}{4}\int_{-\pi/4}^{\pi/4} \frac{dx}{2-\cos 2x}$$ 3. Evaluate $I_1$ using symmetry. Since $\cos 2x$ is an even function, $2-\cos 2x$ is even. Hence $$\frac{x}{2-\cos 2x}$$ is an odd function (odd/even = odd). Therefore, over symmetric limits $[-a,a]$, $$I_1=\int_{-\pi/4}^{\pi/4} \frac{x}{2-\cos 2x}\,dx=0$$ Thus, $$I=I_2=\frac{\pi}{4}\int_{-\pi/4}^{\pi/4} \frac{dx}{2-\cos 2x}$$ 4. Now compute $$J=\int_{-\pi/4}^{\pi/4} \frac{dx}{2-\cos 2x}$$ Use the substitution $$t=\tan x, \qquad dx=\frac{dt}{1+t^2}, \qquad \cos 2x=\frac{1-t^2}{1+t^2}$$ Also, when $x=-\pi/4$, $t=-1$; and when $x=\pi/4$, $t=1$. Then $$2-\cos 2x=2-\frac{1-t^2}{1+t^2} =\frac{2(1+t^2)-(1-t^2)}{1+t^2} =\frac{1+3t^2}{1+t^2}$$ Therefore, $$\frac{dx}{2-\cos 2x} =\frac{\frac{dt}{1+t^2}}{\frac{1+3t^2}{1+t^2}} =\frac{dt}{1+3t^2}$$ So, $$J=\int_{-1}^{1} \frac{dt}{1+3t^2}$$ 5. Evaluate $J$: $$J=\int_{-1}^{1} \frac{dt}{1+3t^2} =\frac{1}{\sqrt{3}}\left[\tan^{-1}(\sqrt{3}t)\right]_{-1}^{1}$$ $$J=\frac{1}{\sqrt{3}}\left(\tan^{-1}(\sqrt{3})-\tan^{-1}(-\sqrt{3})\right)$$ $$J=\frac{1}{\sqrt{3}}\left(\frac{\pi}{3}+\frac{\pi}{3}\right) =\frac{2\pi}{3\sqrt{3}}$$ 6. Hence, $$I=\frac{\pi}{4}\cdot \frac{2\pi}{3\sqrt{3}} =\frac{\pi^2}{6\sqrt{3}}$$ 7. Compare with the options: - A: $\dfrac{\pi^2}{6\sqrt{3}}$ ✅ - B: $\dfrac{\pi^2}{6}$ - C: $\dfrac{\pi^2}{3\sqrt{3}}$ - D: $\dfrac{\pi^2}{12\sqrt{3}}$ Therefore, the correct option is **A**.