The integral integral_0^pi (8x dx) / (4cos^2 x + sin^2 x) is equal to

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q18 – JEE-MAIN 2025

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Home Maths IntegralsQuestion #218

The integral

\int_{0}^{π} \frac{8 x d x}{4 \cos^{2} x + \sin^{2} x}

is equal to

Clue 1 of 3: King's Rule

Apply the property of definite integrals $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ to simplify the numerator.

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Step 1: Apply King's Rule and Simplify✦ Active

Let the given integral be $I$ . First, simplify the denominator using the identity $\sin^{2} x + \cos^{2} x = 1$ : $4 \cos^{2} x + \sin^{2} x = 3 \cos^{2} x + (\cos^{2} x + \sin^{2} x) = 3 \cos^{2} x + 1$ . So, $I = \int_{0}^{π} \frac{8 x d x}{3 \cos^{2} x + 1}$ . Apply the property $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ (King's Rule) with $a = π$ . Since $\cos (π - x) = - \cos x$ , we have $\cos^{2} (π - x) = \cos^{2} x$ . Thus, $I = \int_{0}^{π} \frac{8 (π - x) d x}{3 \cos^{2} x + 1}$ . Adding the original integral and this new form:

2 I = \int_{0}^{π} \frac{8 x + 8 (π - x) d x}{3 \cos^{2} x + 1} = \int_{0}^{π} \frac{8 π d x}{3 \cos^{2} x + 1}

This simplifies to $I = 4 π \int_{0}^{π} \frac{d x}{3 \cos^{2} x + 1}$ . We can rewrite the denominator as $4 \cos^{2} x + \sin^{2} x$ again.

Step 2: Utilize Symmetry and Substitution○ Expand

The integrand $f (x) = \frac{1}{4 \cos^{2} x + \sin^{2} x}$ satisfies $f (π - x) = f (x)$ . Therefore, we can use the property $\int_{0}^{2 a} f (x) d x = 2 \int_{0}^{a} f (x) d x$ for $a = π / 2$ . So, $\int_{0}^{π} \frac{d x}{4 \cos^{2} x + \sin^{2} x} = 2 \int_{0}^{π / 2} \frac{d x}{4 \cos^{2} x + \sin^{2} x}$ . Substituting this back into the expression for $I$ :

I = 4 π (2 \int_{0}^{π / 2} \frac{d x}{4 \cos^{2} x + \sin^{2} x}) = 8 π \int_{0}^{π / 2} \frac{d x}{4 \cos^{2} x + \sin^{2} x}

To evaluate the remaining integral, divide the numerator and denominator by $\cos^{2} x$ :

I = 8 π \int_{0}^{π / 2} \frac{\sec^{2} x d x}{4 + \tan^{2} x}

Now, let $t = \tan x$ . Then $d t = \sec^{2} x d x$ . When $x = 0$ , $t = 0$ . When $x = π / 2$ , $t \to \infty$ . The integral becomes:

I = 8 π \int_{0}^{\infty} \frac{d t}{4 + t^{2}}

💡 Teacher's Secret Hint

Remember to adjust the limits of integration when performing a substitution.

Step 3: Evaluate the Standard Integral○ Expand

The integral is a standard form $\int \frac{d x}{a^{2} + x^{2}} = \frac{1}{a} \arctan (\frac{x}{a})$ . Here, $a^{2} = 4$ , so $a = 2$ . Evaluating the definite integral:

I = 8 π {[\frac{1}{2} \arctan (\frac{t}{2})]}_{0}^{\infty}

Substitute the limits:

I = 8 π (\frac{1}{2} (\arctan (\infty) - \arctan (0)))

Since $\arctan (\infty) = \frac{π}{2}$ and $\arctan (0) = 0$ :

I = 8 π (\frac{1}{2} (\frac{π}{2} - 0)) = 8 π (\frac{π}{4}) = 2 π^{2}

💡 Teacher's Secret Hint

Be careful with the values of $\arctan$ at infinity and zero.

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Maths Question 18 – JEE-MAIN 2025

Clue 1 of 3: King's Rule

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