If {I}=₀^{π/4} (1+ x) d x, then value of I is

Correct Answer is : π8log⁡2π/8 28πlog2

Correct Answer is : π8log⁡2π/8 28πlog2

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MHTCET 11 May 2024 Shift 2 Paper

Question : 115 of 150

Marks: +1, -0

\mathrm{I}=\int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x

, then value of I is

Solution:

Concept:

Use the definite integral property

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

and the identity for

\tan\left(\frac{\pi}{4}-x\right)

Explanation:

Let

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

Using the property, replace

x

\frac{\pi}{4}-x

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

Since

\tan\left(\frac{\pi}{4}-x\right)=\frac{1-\tan x}{1+\tan x}

, we get:

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

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

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

I=\int_{0}^{\frac{\pi}{4}} \log 2\,dx - I

2I=\int_{0}^{\frac{\pi}{4}} \log 2\,dx = \log 2 \cdot \left[x\right]_{0}^{\frac{\pi}{4}}

2I=\frac{\pi}{4}\log 2 \Rightarrow I=\frac{\pi}{8}\log 2

Answer:

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

, which is option C.

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