JEE Mathematics 2023 Question with Solution

Given:

$$I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x + \frac{\pi}{4}}{2 - \cos 2x} \, dx$$

Find: The value of the integral and the correct option.

Using the substitution $x \to -x$,

$$I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{-x + \frac{\pi}{4}}{2 - \cos 2x} \, dx$$

Adding the two forms,

$$2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\pi/2}{2 - \cos 2x} \, dx$$

So,

$$I = \frac{\pi}{4} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2 - \cos 2x} \, dx$$

Since $\cos 2x$ is an even function,

$$I = \frac{\pi}{2} \int_{0}^{\frac{\pi}{4}} \frac{1}{2 - \cos 2x} \, dx$$

Now let $t = \tan x$. Then $dt = \sec^2 x \, dx = (1+t^2) \, dx$ and

$$\cos 2x = \frac{1-t^2}{1+t^2}$$

When $x=0$, $t=0$ and when $x=\frac{\pi}{4}$, $t=1$. Thus,

$$I = \frac{\pi}{2} \int_{0}^{1} \frac{1+t^2}{2(1+t^2)-(1-t^2)} \cdot \frac{dt}{1+t^2}$$ $$I = \frac{\pi}{2} \int_{0}^{1} \frac{1}{3t^2+1} \, dt$$

Let $u=\sqrt{3}t$, so $du=\sqrt{3} \, dt$. The limits become $0$ to $\sqrt{3}$. Therefore,

$$I = \frac{\pi}{2\sqrt{3}} \int_{0}^{\sqrt{3}} \frac{1}{u^2+1} \, du$$ $$I = \frac{\pi}{2\sqrt{3}} \left[ \tan^{-1}(u) \right]_{0}^{\sqrt{3}}$$ $$I = \frac{\pi}{2\sqrt{3}} \left( \frac{\pi}{3} - 0 \right)$$ $$I = \frac{\pi^2}{6\sqrt{3}}$$

Therefore, the value of the integral is $\frac{\pi^2}{6\sqrt{3}}$ and the correct option is D.

Given:

$$I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x + \frac{\pi}{4}}{2 - \cos 2x} \, dx$$

Find: Evaluate the integral using symmetry.

The denominator $2-\cos 2x$ is unchanged under $x \to -x$ because $\cos 2x$ is even. Only the numerator changes from $x+\frac{\pi}{4}$ to $-x+\frac{\pi}{4}$.

So after replacing $x$ by $-x$,

$$I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{-x + \frac{\pi}{4}}{2 - \cos 2x} \, dx$$

Add this to the original expression so that the odd part cancels:

$$2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\left(x + \frac{\pi}{4}\right)+\left(-x + \frac{\pi}{4}\right)}{2 - \cos 2x} \, dx$$ $$2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\pi/2}{2 - \cos 2x} \, dx$$ $$I = \frac{\pi}{4} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{2 - \cos 2x} \, dx$$

Now the integrand is even, hence

$$I = \frac{\pi}{2} \int_{0}^{\frac{\pi}{4}} \frac{1}{2 - \cos 2x} \, dx$$

Use the substitution $t=\tan x$:

$$\cos 2x = \frac{1-t^2}{1+t^2}, \qquad dx = \frac{dt}{1+t^2}$$

Then,

$$2-\cos 2x = 2 - \frac{1-t^2}{1+t^2} = \frac{1+3t^2}{1+t^2}$$

Hence,

$$\frac{1}{2-\cos 2x} \, dx = \frac{1+t^2}{1+3t^2} \cdot \frac{dt}{1+t^2} = \frac{dt}{1+3t^2}$$

Therefore,

$$I = \frac{\pi}{2} \int_{0}^{1} \frac{dt}{1+3t^2}$$

Now substitute $u=\sqrt{3}t$:

$$I = \frac{\pi}{2} \cdot \frac{1}{\sqrt{3}} \int_{0}^{\sqrt{3}} \frac{du}{1+u^2}$$ $$I = \frac{\pi}{2\sqrt{3}} \left[\tan^{-1}u\right]_{0}^{\sqrt{3}} = \frac{\pi}{2\sqrt{3}} \cdot \frac{\pi}{3}$$ $$I = \frac{\pi^2}{6\sqrt{3}}$$

Thus, the correct option is D.