NIMCET 2014 — Mathematics PYQ

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Question

NIMCET 2014 — Mathematics PYQ

NIMCET | Mathematics | 2014

The value of $\int_{0}^{\frac{\pi}{4}} \log(1 + \tan x) dx$ is equal to:

Choose the correct answer:

Explanation

Solution

Concept:

Definite Integral Property: If $\int f(x) dx = g(x) + C$ , then $\int_{a}^{b} f(x) dx = [g(x)]_{a}^{b} = g(b) - g(a)$ .
Another useful property is: $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$ .
Trigonometric Identity: $\tan\left(\frac{\pi}{4} - x\right) = \frac{1 - \tan x}{1 + \tan x}$ .

Calculation:

Let $I = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan x) dx$

Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$ , we get:

\Rightarrow I = \int_{0}^{\frac{\pi}{4}} \log\left[1 + \tan\left(\frac{\pi}{4} - x\right)\right] dx

\Rightarrow I = \int_{0}^{\frac{\pi}{4}} \log\left(1 + \frac{1 - \tan x}{1 + \tan x}\right) dx

\Rightarrow I = \int_{0}^{\frac{\pi}{4}} \log\left(\frac{2}{1 + \tan x}\right) dx

\Rightarrow I = \int_{0}^{\frac{\pi}{4}} (\log 2) dx - \int_{0}^{\frac{\pi}{4}} \log(1 + \tan x) dx

\Rightarrow I = (\log 2) \int_{0}^{\frac{\pi}{4}} 1 dx - I

\Rightarrow 2I = (\log 2) [x]_{0}^{\frac{\pi}{4}}

\Rightarrow 2I = (\log 2) \left(\frac{\pi}{4}\right)

\Rightarrow I = \frac{\pi}{8} \log 2

Correct Option: 3. ( $\frac{\pi}{8} \log 2$ )

Choose the correct answer:

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