NIMCET 2015 — Mathematics PYQ

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Question

NIMCET 2015 — Mathematics PYQ

NIMCET | Mathematics | 2015

The value of $\int_{-\pi/3}^{\pi/3} \frac{x \sin x}{\cos^2 x} dx$ is:

Choose the correct answer:

Explanation

1. Analyze the parity of the function

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

Check if the function is even or odd using $f(-x)$ :

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

Since $f(x) = f(-x)$ , the function is even. Using the property of definite integrals:

\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx

Our integral becomes:

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

2. Integration by Parts

We use the ILATE rule for Integration by Parts: $\int u v dx = u \int v dx - \int (u' \int v dx) dx$ .

Let $u = x$ and $dv = \sec x \tan x dx$ .

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

Applying the formula:

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

3. Evaluate the components

Part 1: $[x \sec x]_{0}^{\pi/3}$

$(\frac{\pi}{3} \sec \frac{\pi}{3}) - (0 \cdot \sec 0) = \frac{\pi}{3} \cdot 2 - 0 = \frac{2\pi}{3}$
Part 2: $\int \sec x dx = \log |\sec x + \tan x|$

$[\log (\sec x + \tan x)]_{0}^{\pi/3} = \log (\sec \frac{\pi}{3} + \tan \frac{\pi}{3}) - \log (\sec 0 + \tan 0)$

$= \log (2 + \sqrt{3}) - \log (1 + 0) = \log (2 + \sqrt{3})$

4. Combine and Simplify

I = 2 \left[ \frac{2\pi}{3} - \log(2 + \sqrt{3}) \right] = \frac{4\pi}{3} - 2 \log(2 + \sqrt{3})

We need to check if $2 + \sqrt{3}$ can be written as $\tan \frac{5\pi}{12}$ .

Note that $\frac{5\pi}{12} = 75^\circ$ .

\tan 75^\circ = \tan(45^\circ + 30^\circ) = \frac{1 + 1/\sqrt{3}}{1 - 1/\sqrt{3}} = \frac{\sqrt{3}+1}{\sqrt{3}-1} = 2 + \sqrt{3}

Thus, $I = \frac{4\pi}{3} - 2 \log \tan \frac{5\pi}{12}$ .

Correct Option: (b)