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MCQMediumJEE 2024Indefinite Integrals

JEE Mathematics 2024 Question with Solution

The integral ∫3/41/4cos⁡(2cot⁡−1(1−x1+x))dx\int_{3/4}^{1/4} \cos\left(2\cot^{-1}\left(\frac{\sqrt{1-x}}{1+x}\right)\right) dx is equal to:

  • A

    −12-\frac{1}{2}

  • B

    14\frac{1}{4}

  • C

    12\frac{1}{2}

  • D

    −14-\frac{1}{4}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given:

I=∫1/43/4cos⁡(2cot⁡−11−x1+x)dxI = \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

θ=cot⁡−11−x1+x\theta = \cot^{-1}\sqrt{\frac{1-x}{1+x}}

Then

cot⁡θ=1−x1+x\cot\theta = \sqrt{\frac{1-x}{1+x}}

Using double-angle identity

From

cot⁡2θ=1−x1+x\cot^2\theta = \frac{1-x}{1+x}

we get

sin⁡2θ=11+cot⁡2θ=11+1−x1+x=1+x2\sin^2\theta = \frac{1}{1+\cot^2\theta} = \frac{1}{1+\frac{1-x}{1+x}} = \frac{1+x}{2}

Now use

cos⁡(2θ)=1−2sin⁡2θ\cos(2\theta) = 1 - 2\sin^2\theta

So,

cos⁡(2θ)=1−2⋅1+x2=−x\cos(2\theta) = 1 - 2\cdot \frac{1+x}{2} = -x

Therefore the integral becomes

I=∫1/43/4−x dx=−[x22]1/43/4I = \int_{1/4}^{3/4} -x \, dx = -\left[\frac{x^2}{2}\right]_{1/4}^{3/4}

Evaluating the limits,

I=−((3/4)22−(1/4)22)=−(932−132)=−832=−14I = -\left(\frac{(3/4)^2}{2} - \frac{(1/4)^2}{2}\right) = -\left(\frac{9}{32} - \frac{1}{32}\right) = -\frac{8}{32} = -\frac{1}{4}

Therefore, the value of the integral is −14-\frac{1}{4}. The correct option is D.

The solution uses limits 14\frac{1}{4} to 34\frac{3}{4}, whereas the given question shows limits 34\frac{3}{4} to 14\frac{1}{4}. The source solution and final answer consistently support −14-\frac{1}{4}, so the answer is taken from the solution as instructed.

Common mistakes

  • Using the wrong identity for cos⁡(2θ)\cos(2\theta) or substituting sin⁡2θ\sin^2\theta and cos⁡2θ\cos^2\theta inconsistently. This gives the wrong sign. First express one trig square correctly in terms of xx, then apply the double-angle formula carefully.

  • Assuming cot⁡θ=1−x1+x\cot\theta = \sqrt{\frac{1-x}{1+x}} directly implies cos⁡θ=1−x1+x\cos\theta = \sqrt{\frac{1-x}{1+x}}. That is incorrect because cotangent is a ratio, not cosine itself. Convert through identities such as sin⁡2θ=11+cot⁡2θ\sin^2\theta = \frac{1}{1+\cot^2\theta} or cos⁡2θ=cot⁡2θ1+cot⁡2θ\cos^2\theta = \frac{\cot^2\theta}{1+\cot^2\theta}.

  • Ignoring the discrepancy in limits between the given question and the solution. Reversing limits changes the sign of a definite integral. Always verify which source is being used before final evaluation.

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