The integral 3/4^1/4 (2 ^-1 ( 1-x1+x)) dx is equal to: — Solution | JEE Mathematics

MCQMediumJEE 2024Definite Integrals

The integral $\int_{3/4}^{1/4} \cos\left(2\cot^{-1}\left(\frac{\sqrt{1-x}}{1+x}\right)\right) \, dx$ is equal to:

Answer

Correct answer:D

Standard Method

Given:

\int_{1/4}^{3/4} \cos\left(2\cot^{-1}\sqrt{\frac{1-x}{1+x}}\right) \, dx

Find: The value of the integral and hence the correct option.

Let

\cot^{-1}\sqrt{\frac{1-x}{1+x}} = \theta

Then

\cot\theta = \sqrt{\frac{1-x}{1+x}}

Using double-angle identity

From

\cot\theta = \sqrt{\frac{1-x}{1+x}}

we get

\sin^2\theta = \frac{1}{1+\cot^2\theta} = \frac{1}{1+\frac{1-x}{1+x}} = \frac{1+x}{2}

Now use the identity

\cos(2\theta) = 1 - 2\sin^2\theta

Therefore,

\cos(2\theta) = 1 - 2\cdot \frac{1+x}{2} = -x

So the integral becomes

\int_{1/4}^{3/4} \cos\left(2\cot^{-1}\sqrt{\frac{1-x}{1+x}}\right) \, dx = \int_{1/4}^{3/4} -x \, dx

Now evaluate:

\int_{1/4}^{3/4} -x \, dx = -\left[\frac{x^2}{2}\right]_{1/4}^{3/4}

= -\left(\frac{\left(\frac{3}{4}\right)^2}{2} - \frac{\left(\frac{1}{4}\right)^2}{2}\right)

= -\left(\frac{9}{32} - \frac{1}{32}\right) = -\frac{8}{32} = -\frac{1}{4}

Therefore, the value of the integral is $-\frac{1}{4}$ . The correct option is D.

Using the limits in the wrong order. The question text shows limits from $\frac{3}{4}$ to $\frac{1}{4}$ , while the extracted solution works with $\frac{1}{4}$ to $\frac{3}{4}$ . Always check the orientation of limits before final evaluation.
Applying the double-angle identity incorrectly. If $\cos(2\theta)$ is not reduced carefully using $\sin^2\theta$ or $\cos^2\theta$ , the integrand will be simplified wrongly. First express one trigonometric square fully in terms of $x$ .
Mistaking $\cot^{-1}$ manipulation. From $\cot\theta = \sqrt{\frac{1-x}{1+x}}$ , students may directly write an incorrect expression for $\sin\theta$ or $\cos\theta$ . Use identities like $\sin^2\theta = \frac{1}{1+\cot^2\theta}$ to avoid sign and algebra errors.

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