JEE MainMathematics

Reviewed by official_key

+4 marks−1 if incorrectNumericalpyq

Integral Calculus: JEE Main Mathematics Question with Solution

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

The target form in the question has as a linear term (not ), which requires squaring the linear expression using the identity .

\int \sqrt{\sec 2x - 1} dx = \alpha \log_e \left| \cos 2x + \beta + \sqrt{\cos 2x \left(1 + \cos \frac{1}{\beta} x\right)} \right| + \text{constant}

, then

\beta - \alpha

is equal to _______.

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

Type the value - units or words beside it are fine.

Step-by-step solutionView

Correct answer

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

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

The integral is $I = \int \sqrt{\sec 2x - 1}\,dx = \alpha \log_e \left| \cos 2x + \beta + \sqrt{\cos 2x \left(1 + \cos \frac{1}{\beta} x\right)} \right| + C$ .

02goal

Find the numerical value of $\beta - \alpha$ .

03approach

Convert $\sec 2x - 1$ to $\frac{1 - \cos 2x}{\cos 2x} = \frac{2\sin^2 x}{\cos 2x}$ . 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: $I = \int \sqrt{\frac{1 - \cos 2x}{\cos 2x}}\,dx = \int \frac{\sqrt{2}\sin x}{\sqrt{2\cos^2 x - 1}}\,dx$ Substitute $t = \sqrt{2}\cos x$ , so $dt = -\sqrt{2}\sin x\,dx$ : $I = -\int \frac{dt}{\sqrt{t^2 - 1}} = -\ln \left| t + \sqrt{t^2 - 1} \right| + C = -\ln \left| \sqrt{2}\cos x + \sqrt{\cos 2x} \right| + C$

05execute

Transform $-\ln |A| = -\frac{1}{2}\ln |A^2|$ : $A^2 = \left(\sqrt{2}\cos x + \sqrt{\cos 2x}\right)^2 = 2\cos^2 x + \cos 2x + 2\sqrt{2}\cos x \sqrt{\cos 2x}$ Substitute $2\cos^2 x = 1 + \cos 2x$ and $\sqrt{2}\cos x = \sqrt{1 + \cos 2x}$ : $A^2 = 2\cos 2x + 1 + 2\sqrt{\cos 2x(1 + \cos 2x)} = 2 \left[ \cos 2x + \frac{1}{2} + \sqrt{\cos 2x(1 + \cos 2x)} \right]$ Absorbing $\ln 2$ into the integration constant: $I = -\frac{1}{2}\ln \left| \cos 2x + \frac{1}{2} + \sqrt{\cos 2x (1 + \cos 2x)} \right| + C'$ Comparing with $\alpha \ln \left| \cos 2x + \beta + \sqrt{\cos 2x\left(1 + \cos \frac{x}{\beta}\right)} \right| + C$ : $\alpha = -\frac{1}{2}, \quad \beta = \frac{1}{2} \implies \frac{1}{\beta} = 2$ Thus, $\beta - \alpha = \frac{1}{2} - \left(-\frac{1}{2}\right) = 1$ .

✓verify

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

Hints that build this answer step by step

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

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 $\cos 2x$ as a linear term (not $\sqrt{2}\cos x$ ), which requires squaring the linear expression $\sqrt{2}\cos x + \sqrt{\cos 2x}$ 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 $\alpha = -\frac{1}{2}$ and $\beta = \frac{1}{2}$ , giving $\beta - \alpha = 1$ .

Why each option works or fails

Step-by-step solution

given: The integral is $I = \int \sqrt{\sec 2x - 1}\,dx = \alpha \log_e \left| \cos 2x + \beta + \sqrt{\cos 2x \left(1 + \cos \frac{1}{\beta} x\right)} \right| + C$ .
goal: Find the numerical value of $\beta - \alpha$ .
approach: Convert $\sec 2x - 1$ to $\frac{1 - \cos 2x}{\cos 2x} = \frac{2\sin^2 x}{\cos 2x}$ . 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: $I = \int \sqrt{\frac{1 - \cos 2x}{\cos 2x}}\,dx = \int \frac{\sqrt{2}\sin x}{\sqrt{2\cos^2 x - 1}}\,dx$ Substitute $t = \sqrt{2}\cos x$ , so $dt = -\sqrt{2}\sin x\,dx$ : $I = -\int \frac{dt}{\sqrt{t^2 - 1}} = -\ln \left| t + \sqrt{t^2 - 1} \right| + C = -\ln \left| \sqrt{2}\cos x + \sqrt{\cos 2x} \right| + C$
execute: Transform $-\ln |A| = -\frac{1}{2}\ln |A^2|$ : $A^2 = \left(\sqrt{2}\cos x + \sqrt{\cos 2x}\right)^2 = 2\cos^2 x + \cos 2x + 2\sqrt{2}\cos x \sqrt{\cos 2x}$ Substitute $2\cos^2 x = 1 + \cos 2x$ and $\sqrt{2}\cos x = \sqrt{1 + \cos 2x}$ : $A^2 = 2\cos 2x + 1 + 2\sqrt{\cos 2x(1 + \cos 2x)} = 2 \left[ \cos 2x + \frac{1}{2} + \sqrt{\cos 2x(1 + \cos 2x)} \right]$ Absorbing $t = \sqrt{2}\cos x$ into the integration constant: $I = -\frac{1}{2}\ln \left| \cos 2x + \frac{1}{2} + \sqrt{\cos 2x (1 + \cos 2x)} \right| + C'$ Comparing with $\alpha \ln \left| \cos 2x + \beta + \sqrt{\cos 2x\left(1 + \cos \frac{x}{\beta}\right)} \right| + C$ : $\alpha = -\frac{1}{2}, \quad \beta = \frac{1}{2} \implies \frac{1}{\beta} = 2$ Thus, $\beta - \alpha = \frac{1}{2} - \left(-\frac{1}{2}\right) = 1$ .
verify: Check that $\frac{1}{\beta} = \frac{1}{1/2} = 2$ , which matches the term $\cos 2x$ inside the square root $\sqrt{\cos 2x(1 + \cos 2x)}$ . Both parameters are completely consistent.

Integral Calculus: JEE Main Mathematics Question with Solution

What feels right?

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

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

We think you should solve this next

The value of $\int_{0}^{20\pi} (\sin^4 x + \cos^4 x)dx$ is equal to…

Let…

If the area of the region…

Students also ask

Integral Calculus: JEE Main Mathematics Question with Solution

By substituting u=cos⁡xu = \cos xu=cosx and rewriting the argument of the logarithm, we find α=−12\alpha = -\frac{1}{2}α=−21​ and β=12\beta = \frac{1}{2}β=21​, giving β−α=1\beta - \alpha = 1β−α=1.

01given

02goal

03approach

04execute

05execute

✓verify

The value of ∫020π(sin⁡4x+cos⁡4x)dx\int_{0}^{20\pi} (\sin^4 x + \cos^4 x)dx∫020π​(sin4x+cos4x)dx is equal to…

Let…

If the area of the region…

Students also ask

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

The value of $\int_{0}^{20\pi} (\sin^4 x + \cos^4 x)dx$ is equal to…