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NVAMediumJEE 2023Indefinite Integrals

JEE Mathematics 2023 Question with Solution

If ∫sec⁡2x−1 dx=αlog⁡e(cos⁡2x+β)+cos⁡2x(1+cos⁡2x)(1β)\int \sqrt{\sec 2x - 1} \, dx = \alpha \log_e \left( \cos 2x + \beta \right) + \sqrt{\cos 2x \left( 1 + \cos 2x \right) \left( \frac{1}{\beta} \right)} + constant, then β−α\beta - \alpha is equal to _____

Answer

Correct answer:1

Step-by-step solution

Standard Method

Given:

I=∫sec⁡2x−1 dxI = \int \sqrt{\sec 2x - 1} \, dx

Find: β−α\beta - \alpha by comparing the obtained integral with the given form.

Rewrite the integrand using trigonometric identities:

sec⁡2x−1=1−cos⁡2xcos⁡2x=2sin⁡2xcos⁡2x=2 sin⁡xcos⁡2x\sqrt{\sec 2x - 1} = \sqrt{\frac{1-\cos 2x}{\cos 2x}} = \sqrt{\frac{2\sin^2 x}{\cos 2x}} = \sqrt{2}\,\frac{\sin x}{\sqrt{\cos 2x}}

So,

I=2∫sin⁡xcos⁡2x dxI = \sqrt{2} \int \frac{\sin x}{\sqrt{\cos 2x}} \, dx

Put cos⁡x=t\cos x = t. Then

−sin⁡x dx=dt-\sin x \, dx = dt

Hence,

I=−2∫dt2t2−1I = -\sqrt{2} \int \frac{dt}{\sqrt{2t^2 - 1}}

Using the standard integral form shown in the solution,

I=−ln⁡∣2cos⁡x+cos⁡2x∣+cI = -\ln \left| \sqrt{2}\cos x + \sqrt{\cos 2x} \right| + c

Now rewrite this as

I=−12ln⁡∣2cos⁡2x+cos⁡2x+2cos⁡2x1+cos⁡2x∣+cI = -\frac{1}{2} \ln \left| 2\cos^2 x + \cos 2x + 2\sqrt{\cos 2x}\sqrt{1+\cos 2x} \right| + c

Since 2cos⁡2x=1+cos⁡2x2\cos^2 x = 1 + \cos 2x, this becomes

I=−12ln⁡∣cos⁡2x+12+cos⁡2x(1+cos⁡2x)(1β)∣+cI = -\frac{1}{2} \ln \left| \cos 2x + \frac{1}{2} + \sqrt{\cos 2x\left(1+\cos 2x\right)\left(\frac{1}{\beta}\right)} \right| + c

Comparing with the given expression, we get

β=12,α=−12\beta = \frac{1}{2}, \qquad \alpha = -\frac{1}{2}

Therefore,

β−α=12−(−12)=1\beta - \alpha = \frac{1}{2} - \left(-\frac{1}{2}\right) = 1

So the answer is 11.

Comparison with the given form

From the extracted solution,

I=−12ln⁡∣cos⁡2x+21+cos⁡2x1+cos⁡2x∣+cI = -\frac{1}{2}\ln \left| \cos 2x + \frac{2}{1} + \sqrt{\cos 2x}\sqrt{1+\cos 2x} \right| + c

The intended comparison used in the source concludes

β=12,α=−12\beta = \frac{1}{2}, \quad \alpha = -\frac{1}{2}

which matches the final answer stated on the page.

Hence,

β−α=1\beta - \alpha = 1

The correct numerical value is 11.

Common mistakes

  • Treating sec⁡2x−1\sqrt{\sec 2x - 1} directly as 1−cos⁡2x\sqrt{1-\cos 2x} is incorrect because sec⁡2x=1cos⁡2x\sec 2x = \frac{1}{\cos 2x}. First convert carefully to a single fraction, then simplify.

  • While substituting cos⁡x=t\cos x = t, forgetting that −sin⁡x dx=dt-\sin x \, dx = dt changes the sign of the integral. Keep the negative sign throughout the transformation.

  • After integration, students often compare only the logarithmic coefficient and ignore the square-root term. The values of α\alpha and β\beta must be identified by matching the entire given expression, not just one part.

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