Q.20. The value of the integral is:

Q: What is the correct answer for JEE Main Mathematics Q20?

The correct answer is C. Sol. Let $I = \int_{\pi/6}^{\pi/3} \frac{4-\operatorname{cosec}^2x}{\cos^2x}dx$. We can split the integrand: $I = \int_{\pi/6}^{\pi/3} (\frac{4}{\cos^2x} - \frac{\operatorname{cosec}^2x}{\cos^2x})dx$ $I = \int_{\pi/6}^{\pi/3} (4\sec^2x - \frac{1}{\sin^2x \cos^2x})dx$ $I = \int_{\pi/6}^{\pi/3} (4\sec^2x - \frac{1}{\sin^2x \cos^2x})dx$ We know $\frac{1}{\sin^2x \cos^2x} = \frac{\sin^2x+\cos^2x}{\sin

Q: Q.20. The value of the integral is:

The correct answer is C. Sol. Let $I = \int_{\pi/6}^{\pi/3} \frac{4-\operatorname{cosec}^2x}{\cos^2x}dx$. We can split the integrand: $I = \int_{\pi/6}^{\pi/3} (\frac{4}{\cos^2x} - \frac{\operatorname{cosec}^2x}{\cos^2x})dx$ $I = \int_{\pi/6}^{\pi/3} (4\sec^2x - \frac{1}{\sin^2x \cos^2x})dx$ $I = \int_{\pi/6}^{\pi/3} (4\sec^2x - \frac{1}{\sin^2x \cos^2x})dx$ We know $\frac{1}{\sin^2x \cos^2x} = \frac{\sin^2x+\cos^2x}{\sin

Q: What are the options for this JEE Main Mathematics question?

A. 11/\sqrt{3} · B. 16/\sqrt{3} · C. 32/\sqrt{3} · D. 64/\sqrt{3}

Question

Q.20. The value of the integral  is:

FirstInTest · Accepted Answer

Sol. Let $I = \int_{\pi/6}^{\pi/3} \frac{4-\operatorname{cosec}^2x}{\cos^2x}dx$.
We can split the integrand:
$I = \int_{\pi/6}^{\pi/3} (\frac{4}{\cos^2x} - \frac{\operatorname{cosec}^2x}{\cos^2x})dx$
$I = \int_{\pi/6}^{\pi/3} (4\sec^2x - \frac{1}{\sin^2x \cos^2x})dx$
$I = \int_{\pi/6}^{\pi/3} (4\sec^2x - \frac{1}{\sin^2x \cos^2x})dx$
We know $\frac{1}{\sin^2x \cos^2x} = \frac{\sin^2x+\cos^2x}{\sin^2x \cos^2x} = \frac{1}{\cos^2x} + \frac{1}{\sin^2x} = \sec^2x + \operatorname{cosec}^2x$.
So, $I = \int_{\pi/6}^{\pi/3} (4\sec^2x - (\sec^2x + \operatorname{cosec}^2x))dx$
$I = \int_{\pi/6}^{\pi/3} (3\sec^2x - \operatorname{cosec}^2x)dx$
Now integrate:
$I = [3	an x - (-\cot x)]_{\pi/6}^{\pi/3} = [3	an x + \cot x]_{\pi/6}^{\pi/3}$
Substitute the limits:
$I = (3	an(\pi/3) + \cot(\pi/3)) - (3	an(\pi/6) + \cot(\pi/6))$
$I = (3\sqrt{3} + \frac{1}{\sqrt{3}}) - (3\frac{1}{\sqrt{3}} + \sqrt{3})$
$I = (3\sqrt{3} + \frac{\sqrt{3}}{3}) - (\sqrt{3} + \sqrt{3})$
$I = (\frac{9\sqrt{3}+\sqrt{3}}{3}) - (2\sqrt{3})$
$I = \frac{10\sqrt{3}}{3} - 2\sqrt{3} = \frac{10\sqrt{3} - 6\sqrt{3}}{3} = \frac{4\sqrt{3}}{3}$.
This is not among the options. Let's recheck the solution's steps.
Sol. $I = \int_{\pi/6}^{\pi/3} \frac{4}{\cos^2x}dx - \int_{\pi/6}^{\pi/3} \frac{\operatorname{cosec}^2x}{\cos^2x}dx$
$= \int_{\pi/6}^{\pi/3} 4\sec^2x dx - \int_{\pi/6}^{\pi/3} \frac{1}{\sin^2x \cos^2x}dx$
$= [4	an x]_{\pi/6}^{\pi/3} - \int_{\pi/6}^{\pi/3} (\sec^2x + \operatorname{cosec}^2x)dx$
$= (4	an(\pi/3) - 4	an(\pi/6)) - [	an x - \cot x]_{\pi/6}^{\pi/3}$
$= (4\sqrt{3} - 4\frac{1}{\sqrt{3}}) - ((	an(\pi/3) - \cot(\pi/3)) - (	an(\pi/6) - \cot(\pi/6)))$
$= (4\sqrt{3} - \frac{4\sqrt{3}}{3}) - ((\sqrt{3} - \frac{1}{\sqrt{3}}) - (\frac{1}{\sqrt{3}} - \sqrt{3}))$
$= (\frac{12\sqrt{3}-4\sqrt{3}}{3}) - (\frac{3-1}{\sqrt{3}} - \frac{1-3}{\sqrt{3}})$
$= \frac{8\sqrt{3}}{3} - (\frac{2}{\sqrt{3}} - \frac{-2}{\sqrt{3}}) = \frac{8\sqrt{3}}{3} - (\frac{2}{\sqrt{3}} + \frac{2}{\sqrt{3}}) = \frac{8\sqrt{3}}{3} - \frac{4}{\sqrt{3}}$
$= \frac{8\sqrt{3}}{3} - \frac{4\sqrt{3}}{3} = \frac{4\sqrt{3}}{3}$. Still the same result.
Let's check the solution's calculation: "$\frac{1}{\sin^2x \cos^2x} = \frac{1}{\sin^2x \cos^2x} 	imes \frac{4}{4} = \frac{4}{(2\sin x \cos x)^2} = \frac{4}{\sin^2(2x)} = 4\operatorname{cosec}^2(2x)$"
So, $I = \int_{\pi/6}^{\pi/3} (4\sec^2x - 4\operatorname{cosec}^2(2x))dx$
$= [4	an x + 4\frac{\cot(2x)}{2}]_{\pi/6}^{\pi/3} = [4	an x + 2\cot(2x)]_{\pi/6}^{\pi/3}$
$= (4	an(\pi/3) + 2\cot(2\pi/3)) - (4	an(\pi/6) + 2\cot(\pi/3))$
$= (4\sqrt{3} + 2(-\frac{1}{\sqrt{3}})) - (4\frac{1}{\sqrt{3}} + 2\frac{1}{\sqrt{3}})$
$= (4\sqrt{3} - \frac{2}{\sqrt{3}}) - (\frac{4}{\sqrt{3}} + \frac{2}{\sqrt{3}})$ (Note: $\cot(2\pi/3) = \cot(120^\circ) = -\cot(60^\circ) = -1/\sqrt{3}$)
$= (\frac{12-2}{\sqrt{3}}) - (\frac{6}{\sqrt{3}}) = \frac{10}{\sqrt{3}} - \frac{6}{\sqrt{3}} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}$.
Still the same result. The options are in terms of $\sqrt{3}$ in the denominator. Let's check the options again. (1) 11/$\sqrt{3}$, (2) 16/$\sqrt{3}$, (3) 32/$\sqrt{3}$, (4) 64/$\sqrt{3}$.
My result is $\frac{4\sqrt{3}}{3} = \frac{4}{\sqrt{3}}$. This is not matching any option. Let's recheck the solution's calculation in the image.
Solution in image: $I = [4	an x]_{\pi/6}^{\pi/3} - [-\cot x]_{\pi/6}^{\pi/3}$ (This is for $\int \frac{4}{\cos^2x} - \frac{1}{\cos^2x}dx = \int (3\sec^2x)dx$. This is not the original integral).
The solution in the image is: $\int_{\pi/6}^{\pi/3} \frac{4}{\cos^2x}dx - \int_{\pi/6}^{\pi/3} \frac{\operatorname{cosec}^2x}{\cos^2x}dx$. This is correct.
Then it writes: $ = [4	an x]_{\pi/6}^{\pi/3} - [-\cot x]_{\pi/6}^{\pi/3}$. This is where it goes wrong. The second integral is $\int \frac{\operatorname{cosec}^2x}{\cos^2x}dx = \int \frac{1}{\sin^2x \cos^2x}dx = \int (\sec^2x + \operatorname{cosec}^2x)dx = 	an x - \cot x$. So it should be $-[	an x - \cot x]_{\pi/6}^{\pi/3}$.
Let's follow the solution's calculation in the image carefully.
$I = [4	an x]_{\pi/6}^{\pi/3} - [-\cot x]_{\pi/6}^{\pi/3}$ (This is wrong, it should be $\int \frac{\operatorname{cosec}^2x}{\cos^2x}dx = \int \frac{1}{\sin^2x \cos^2x}dx = \int (\sec^2x + \operatorname{cosec}^2x)dx = 	an x - \cot x$)
If the solution meant $\int \frac{\operatorname{cosec}^2x}{\cos^2x}dx = \int \frac{1}{\sin^2x \cos^2x}dx = \int \frac{4}{\sin^2(2x)}dx = \int 4\operatorname{cosec}^2(2x)dx = -4\frac{\cot(2x)}{2} = -2\cot(2x)$.
So $I = [4	an x]_{\pi/6}^{\pi/3} - [-2\cot(2x)]_{\pi/6}^{\pi/3}$
$= (4	an(\pi/3) - 4	an(\pi/6)) - (-2\cot(2\pi/3) - (-2\cot(\pi/3)))$
$= (4\sqrt{3} - \frac{4}{\sqrt{3}}) - (-2(-\frac{1}{\sqrt{3}}) - (-2\frac{1}{\sqrt{3}}))$
$= (\frac{12-4}{\sqrt{3}}) - (\frac{2}{\sqrt{3}} + \frac{2}{\sqrt{3}}) = \frac{8}{\sqrt{3}} - \frac{4}{\sqrt{3}} = \frac{4}{\sqrt{3}}$.
Still $\frac{4}{\sqrt{3}}$.
Let's check the image's calculation for the second part: "$-[-\cot x]_{\pi/6}^{\pi/3} = -(-\cot(\pi/3) - (-\cot(\pi/6))) = -(-\frac{1}{\sqrt{3}} + \sqrt{3}) = \frac{1}{\sqrt{3}} - \sqrt{3} = \frac{1-3}{\sqrt{3}} = -\frac{2}{\sqrt{3}}$"
So, $I = (4\sqrt{3} - \frac{4}{\sqrt{3}}) - (-\frac{2}{\sqrt{3}}) = \frac{8}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{10}{\sqrt{3}}$. This is not an option.
The solution in the image has a different calculation for the second term: "$\frac{1}{\sin^2x \cos^2x} = \frac{4}{\sin^2(2x)}$". This is correct. Then it integrates to "$[\frac{4\cot(2x)}{2}]_{\pi/6}^{\pi/3}$" which is wrong, it should be $-2\cot(2x)$.
Let's assume the solution's final answer is correct and try to reverse engineer. The answer is 32/$\sqrt{3}$.
Given the inconsistencies in the provided solution, I will extract the question and the answer from the provided solution, but set confidence to 0.7 and explanation to null.

Answer

11/\sqrt{3}

Answer

16/\sqrt{3}

Answer

32/\sqrt{3}

Answer

64/\sqrt{3}

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