JEE MainMathematics

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+4 marks−1 if incorrectNumericalPrevious-year question

Indefinite Integration of a Trigonometric Function with Square Roots

Q: Why did we square the argument and put a factor of 1/2 outside the logarithm?

The target form in the question has cos 2x as a linear term (not √(2)cos x), which requires squaring the linear expression √(2)cos x + √(cos 2x) using the identity ln |u| = 1/2ln |u^2|.

If integral of √(sec 2x - 1) dx = α log_e | cos 2x + β + √(cos 2x (1 + cos 1/β x)) | + constant, then β - α is equal to _______.

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Step-by-step solutionView

Correct answer

By substituting $u = \cos x$ and rewriting the argument of the logarithm, we find $α = - \frac{1}{2}$ and $β = \frac{1}{2}$ , giving $β - α = 1$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01Given

The integral is $I = \int \sqrt{\sec 2 x - 1} d x = α \log_{e} ∣ \cos 2 x + β + \sqrt{\cos 2 x (1 + \cos \frac{1}{β} x)} ∣ + C$ .

02Goal

Find the numerical value of $β - α$ .

03Approach

Convert $\sec 2 x - 1$ to $\frac{1 - \cos 2 x}{\cos 2 x} = \frac{2 \sin^{2} x}{\cos 2 x}$ . Substitute $t = \sqrt{2} \cos x$ to integrate using standard forms, then manipulate the argument of the logarithm into the target form.

04Execute

Rewrite the integrand: $I = \int \sqrt{\frac{1 - \cos 2 x}{\cos 2 x}} d x = \int \frac{\sqrt{2} \sin x}{\sqrt{2 \cos^{2} x - 1}} d x$ Substitute $t = \sqrt{2} \cos x$ , so $d t = - \sqrt{2} \sin x d x$ : $I = - \int \frac{d t}{\sqrt{t^{2} - 1}} = - \ln ∣ t + \sqrt{t^{2} - 1} ∣ + C = - \ln ∣ \sqrt{2} \cos x + \sqrt{\cos 2 x} ∣ + C$

05Execute

Transform $- \ln ∣ A ∣ = - \frac{1}{2} \ln ∣ A^{2} ∣$ : $A^{2} = {(\sqrt{2} \cos x + \sqrt{\cos 2 x})}^{2} = 2 \cos^{2} x + \cos 2 x + 2 \sqrt{2} \cos x \sqrt{\cos 2 x}$ Substitute $2 \cos^{2} x = 1 + \cos 2 x$ and $\sqrt{2} \cos x = \sqrt{1 + \cos 2 x}$ : $A^{2} = 2 \cos 2 x + 1 + 2 \sqrt{\cos 2 x (1 + \cos 2 x)} = 2 [\cos 2 x + \frac{1}{2} + \sqrt{\cos 2 x (1 + \cos 2 x)}]$ Absorbing $\ln 2$ into the integration constant: $I = - \frac{1}{2} \ln ∣ \cos 2 x + \frac{1}{2} + \sqrt{\cos 2 x (1 + \cos 2 x)} ∣ + C^{'}$ Comparing with $α \ln ∣ \cos 2 x + β + \sqrt{\cos 2 x (1 + \cos \frac{x}{β})} ∣ + C$ : $α = - \frac{1}{2}, β = \frac{1}{2} ⟹ \frac{1}{β} = 2$ Thus, $β - α = \frac{1}{2} - (- \frac{1}{2}) = 1$ .

✓Verify

Check that $\frac{1}{β} = \frac{1}{1 / 2} = 2$ , which matches the term $\cos 2 x$ inside the square root $\sqrt{\cos 2 x (1 + \cos 2 x)}$ . Both parameters are completely consistent.

Hints that build this answer step by step

What is the most effective way to rewrite $\sqrt{\sec 2 x - 1}$ into a form suitable for substitution?
Express $\sec 2 x$ as $\frac{1}{\cos 2 x}$ , giving $\sqrt{\frac{1 - \cos 2 x}{\cos 2 x}} = \frac{\sqrt{2} \sin x}{\sqrt{\cos 2 x}}$ , and substitute $u = \cos x$
With $u = \cos x$ , where $\cos 2 x = 2 u^{2} - 1$ and $d u = - \sin x d x$ , what does the integral evaluate to in terms of $u$ ?
$- \log_{e} ∣ u + \sqrt{u^{2} - \frac{1}{2}} ∣ + C$
Using the logarithmic identity $- \log_{e} ∣ A ∣ = - \frac{1}{2} \log_{e} ∣ A^{2} ∣$ , what is ${(u + \sqrt{u^{2} - \frac{1}{2}})}^{2}$ in terms of $\cos 2 x$ ?
$\cos 2 x + \frac{1}{2} + \sqrt{\cos 2 x (1 + \cos 2 x)}$
Comparing $- \frac{1}{2} \log_{e} ∣ \cos 2 x + \frac{1}{2} + \sqrt{\cos 2 x (1 + \cos 2 x)} ∣$ to the given stem, what are $α$ , $β$ , and $β - α$ ?
$α = - \frac{1}{2}$ , $β = \frac{1}{2}$ , so $β - α = 1$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

Quick checks

Students also ask

Why did we square the argument and put a factor of 1/2 outside the logarithm?

The target form in the question has $\cos 2 x$ as a linear term (not $\sqrt{2} \cos x$ ), which requires squaring the linear expression $\sqrt{2} \cos x + \sqrt{\cos 2 x}$ using the identity $\ln ∣ u ∣ = \frac{1}{2} \ln ∣ u^{2} ∣$ .

Answer

By substituting $u = \cos x$ and rewriting the argument of the logarithm, we find $α = - \frac{1}{2}$ and $β = \frac{1}{2}$ , giving $β - α = 1$ .

Why each option works or fails

Step-by-step solution

Given: The integral is $I = \int \sqrt{\sec 2 x - 1} d x = α \log_{e} ∣ \cos 2 x + β + \sqrt{\cos 2 x (1 + \cos \frac{1}{β} x)} ∣ + C$ .
Goal: Find the numerical value of $β - α$ .
Approach: Convert $\sec 2 x - 1$ to $\frac{1 - \cos 2 x}{\cos 2 x} = \frac{2 \sin^{2} x}{\cos 2 x}$ . Substitute $t = \sqrt{2} \cos x$ to integrate using standard forms, then manipulate the argument of the logarithm into the target form.
Execute: Rewrite the integrand: $I = \int \sqrt{\frac{1 - \cos 2 x}{\cos 2 x}} d x = \int \frac{\sqrt{2} \sin x}{\sqrt{2 \cos^{2} x - 1}} d x$ Substitute $t = \sqrt{2} \cos x$ , so $d t = - \sqrt{2} \sin x d x$ : $I = - \int \frac{d t}{\sqrt{t^{2} - 1}} = - \ln ∣ t + \sqrt{t^{2} - 1} ∣ + C = - \ln ∣ \sqrt{2} \cos x + \sqrt{\cos 2 x} ∣ + C$
Execute: Transform $- \ln ∣ A ∣ = - \frac{1}{2} \ln ∣ A^{2} ∣$ : $A^{2} = {(\sqrt{2} \cos x + \sqrt{\cos 2 x})}^{2} = 2 \cos^{2} x + \cos 2 x + 2 \sqrt{2} \cos x \sqrt{\cos 2 x}$ Substitute $2 \cos^{2} x = 1 + \cos 2 x$ and $\sqrt{2} \cos x = \sqrt{1 + \cos 2 x}$ : $A^{2} = 2 \cos 2 x + 1 + 2 \sqrt{\cos 2 x (1 + \cos 2 x)} = 2 [\cos 2 x + \frac{1}{2} + \sqrt{\cos 2 x (1 + \cos 2 x)}]$ Absorbing $\ln 2$ into the integration constant: $I = - \frac{1}{2} \ln ∣ \cos 2 x + \frac{1}{2} + \sqrt{\cos 2 x (1 + \cos 2 x)} ∣ + C^{'}$ Comparing with $α \ln ∣ \cos 2 x + β + \sqrt{\cos 2 x (1 + \cos \frac{x}{β})} ∣ + C$ : $α = - \frac{1}{2}, β = \frac{1}{2} ⟹ \frac{1}{β} = 2$ Thus, $β - α = \frac{1}{2} - (- \frac{1}{2}) = 1$ .
Verify: Check that $\frac{1}{β} = \frac{1}{1 / 2} = 2$ , which matches the term $\cos 2 x$ inside the square root $\sqrt{\cos 2 x (1 + \cos 2 x)}$ . Both parameters are completely consistent.

Indefinite Integration of a Trigonometric Function with Square Roots

What feels right?

By substituting $u = \cos x$ and rewriting the argument of the logarithm, we find $α = - \frac{1}{2}$ and $β = \frac{1}{2}$ , giving $β - α = 1$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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