If _{π / 6}^{π / 3} 1/1+ x+ x d x= 2, then the value of _{π / 6}^{π / 3} x/1+ x+ x d x= \_\_\_\_

Correct Answer is : π12−12log⁡2π/12-1/2 212π−21log2

Correct Answer is : π12−12log⁡2π/12-1/2 212π−21log2

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MHTCET 13 Apr 2026 Shift 2 Paper

Question : 132 of 150

Marks: +1, -0

\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\sin x+\cos x} d x=\log 2

\int_{\pi / 6}^{\pi / 3} \frac{\cos x}{1+\sin x+\cos x} d x=

\_\_\_\_

Solution:

Concept:

Use the definite integral property

\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx

to relate the two integrals.

Explanation:

Let

I = \int_{\pi/6}^{\pi/3} \frac{\cos x}{1+\sin x+\cos x}\,dx

Here,

a = \frac{\pi}{6}

and

b = \frac{\pi}{3}

, so

a+b = \frac{\pi}{2}

Applying the property:

I = \int_{\pi/6}^{\pi/3} \frac{\cos\left(\frac{\pi}{2}-x\right)}{1+\sin\left(\frac{\pi}{2}-x\right)+\cos\left(\frac{\pi}{2}-x\right)}\,dx

Using

\cos\left(\frac{\pi}{2}-x\right)=\sin x

and

\sin\left(\frac{\pi}{2}-x\right)=\cos x

, we get:

I = \int_{\pi/6}^{\pi/3} \frac{\sin x}{1+\sin x+\cos x}\,dx

Now add the given integral,

\int_{\pi/6}^{\pi/3} \frac{1}{1+\sin x+\cos x}\,dx = \log 2

, to both

I

expressions:

\log 2 + I + I = \int_{\pi/6}^{\pi/3} \frac{1+\sin x+\cos x}{1+\sin x+\cos x}\,dx

This simplifies to:

\log 2 + 2I = \int_{\pi/6}^{\pi/3} 1\,dx

Therefore,

\log 2 + 2I = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}

2I = \frac{\pi}{6} - \log 2

I = \frac{\pi}{12} - \frac{1}{2}\log 2

Answer:

\frac{\pi}{12} - \frac{1}{2}\log 2

Correct option: A

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