_{π/3}² π/3} x/1+ x d x=

Correct Answer is : π(2−3)π(2-{3})π(2−3)

Correct Answer is : π(2−3)π(2-{3})π(2−3)

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MHTCET 22 Apr 2025 Shift 1 Paper

Question : 128 of 150

Marks: +1, -0

\int_{\frac{\pi}{3}}^{\frac{2 \pi}{3}} \frac{x}{1+\sin x} d x=

Solution:

Concept:

Use the definite integral property

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

, then add the two forms to simplify.

Explanation:

Let

\mathrm{I} = \int_{\pi/3}^{2\pi/3} \frac{x}{1+\sin x} dx

Using the property with

a = \frac{\pi}{3}

and

b = \frac{2\pi}{3}

, we have

a+b = \pi

So,

x

is replaced by

\pi - x

and

\sin(\pi - x) = \sin x

Thus,

\mathrm{I} = \int_{\pi/3}^{2\pi/3} \frac{\pi - x}{1+\sin x} dx

Adding the two expressions:

2\mathrm{I} = \int_{\pi/3}^{2\pi/3} \frac{x + (\pi - x)}{1+\sin x} dx = \pi \int_{\pi/3}^{2\pi/3} \frac{1}{1+\sin x} dx

Rationalize:

\frac{1}{1+\sin x} \times \frac{1-\sin x}{1-\sin x} = \frac{1-\sin x}{\cos^2 x} = \sec^2 x - \sec x \tan x

So,

2\mathrm{I} = \pi \int_{\pi/3}^{2\pi/3} (\sec^2 x - \sec x \tan x) dx

2\mathrm{I} = \pi \left[ \tan x - \sec x \right]_{\pi/3}^{2\pi/3}

Evaluate:

\tan\frac{2\pi}{3} - \sec\frac{2\pi}{3} = -\sqrt{3} - (-2) = 2-\sqrt{3}

Also,

\tan\frac{\pi}{3} - \sec\frac{\pi}{3} = \sqrt{3} - 2

Therefore,

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

Hence,

\mathrm{I} = \pi(2-\sqrt{3})

Answer:

\pi(2-\sqrt{3})

i.e., option B.

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