NIMCET 2023 — Mathematics PYQ

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Question

NIMCET 2023 — Mathematics PYQ

NIMCET | Mathematics | 2023

The value of $\int_{- π /3}^{π /3} \frac{x s i n x}{c o s ^{2} x} d x$ is

Choose the correct answer:

Explanation

Step 1: Identify the property of the definite integral

The integral is in the form $\int_{-a}^{a} f(x) \, dx$ . We check if the integrand $f(x)$ is even or odd.

If $f(-x) = f(x)$ , the function is even, and $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ .
If $f(-x) = -f(x)$ , the function is odd, and $\int_{-a}^{a} f(x) \, dx = 0$ .

Let $f(x) = \frac{x \sin x}{\cos^2 x}$ .

f(-x) = \frac{(-x) \sin(-x)}{\cos^2(-x)}

Since $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$ :

f(-x) = \frac{(-x)(-\sin x)}{(\cos x)^2} = \frac{x \sin x}{\cos^2 x} = f(x)

Thus, $f(x)$ is an even function.

Step 2: Simplify the integral

I = 2 \int_{0}^{\pi/3} \frac{x \sin x}{\cos^2 x} \, dx = 2 \int_{0}^{\pi/3} x \cdot (\sec x \tan x) \, dx

Step 3: Integrate by parts

Let $u = x$ (First function) and $dv = \sec x \tan x \, dx$ (Second function).

Then $du = dx$ and $v = \sec x$ .

Using the formula $\int u \, dv = uv - \int v \, du$ :

I = 2 \left[ [x \sec x]_{0}^{\pi/3} - \int_{0}^{\pi/3} \sec x \, dx \right]

Step 4: Evaluate the limits and the remaining integral

The integral of $\sec x$ is $\log_e|\sec x + \tan x|$ .

I = 2 \left[ \left( \frac{\pi}{3} \sec \frac{\pi}{3} - 0 \right) - [\log_e(\sec x + \tan x)]_{0}^{\pi/3} \right]

Substitute values ( $\sec \frac{\pi}{3} = 2$ and $\tan \frac{\pi}{3} = \sqrt{3}$ ):

I = 2 \left[ \left( \frac{\pi}{3} \cdot 2 \right) - \left( \log_e(2 + \sqrt{3}) - \log_e(1 + 0) \right) \right]

I = 2 \left[ \frac{2\pi}{3} - \log_e(2 + \sqrt{3}) \right]

I = \frac{4\pi}{3} - 2\log_e(2 + \sqrt{3})

Correct Option:

(B) $\frac{4\pi}{3} - 2\log_e(2 + \sqrt{3})$