₀^{π/4} ≤ft( x+ x/ x) d x=

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

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

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MHTCET 10 May 2024 Shift 2 Paper

Question : 107 of 150

Marks: +1, -0

\int_{0}^{\frac{\pi}{4}} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x=

Solution:

Concept:

Use the definite integral property

\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx

to simplify the logarithmic integrand.

Explanation:

Let

I = \int_{0}^{\frac{\pi}{4}} \log\left(\frac{\sin x+\cos x}{\cos x}\right) dx

Since

\frac{\sin x+\cos x}{\cos x} = 1+\tan x

, we get:

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

Apply the property

\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx

with

a = \frac{\pi}{4}

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

Using

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

1+\tan\left(\frac{\pi}{4}-x\right) = 1+\frac{1-\tan x}{1+\tan x} = \frac{2}{1+\tan x}

Thus:

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

The second integral is again

I

, so:

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

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

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

Answer:

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

Correct option: D.

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

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