The integral ∫₀^π (x+3)sin x / (1+3cos² x) dx is equal to

The correct answer is Option 4. 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 #222

The integral

\int_{0}^{π} \frac{(x + 3) \sin x}{1 + 3 \cos^{2} x} d x

is equal to

Clue 1 of 3: King's Property for Definite Integrals

When the limits of integration are $0$ to $a$ or $a$ to $b$ , consider using the property $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ .

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Ninja Strategy•King's Property Factor Analysis

Recognize that applying King's property to the integral will result in a factor of $(π + 6)$ in the numerator, which immediately points to the correct option.

🧠 Full Solution Path

Step 1: Apply King's Property✦ Active

Let the given integral be $I$ . Apply the property $\int_{0}^{π} f (x) d x = \int_{0}^{π} f (π - x) d x$ to the integral.

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

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

💡 Teacher's Secret Hint

Remember that $\sin (π - x) = \sin x$ and $\cos^{2} (π - x) = \cos^{2} x$ .

Step 2: Combine and Simplify○ Expand

Add the original integral and the transformed integral. This eliminates the $x$ term from the numerator.

2 I = \int_{0}^{π} \frac{(x + 3 + π - x + 3) \sin x}{1 + 3 \cos^{2} x} d x = \int_{0}^{π} \frac{(π + 6) \sin x}{1 + 3 \cos^{2} x} d x

2 I = (π + 6) \int_{0}^{π} \frac{\sin x}{1 + 3 \cos^{2} x} d x

💡 Teacher's Secret Hint

This step is key to simplifying the integrand significantly.

Step 3: Evaluate the Remaining Integral○ Expand

Substitute $t = \cos x$ , so $d t = - \sin x d x$ . Change the limits of integration from $x = 0$ to $t = 1$ and $x = π$ to $t = - 1$ . Use the property $\int_{a}^{b} f (t) d t = - \int_{b}^{a} f (t) d t$ and for even functions $\int_{- a}^{a} f (t) d t = 2 \int_{0}^{a} f (t) d t$ .

2 I = (π + 6) \int_{1}^{- 1} \frac{- d t}{1 + 3 t^{2}} = (π + 6) \int_{- 1}^{1} \frac{d t}{1 + 3 t^{2}}

2 I = (π + 6) \cdot 2 \int_{0}^{1} \frac{d t}{1 + 3 t^{2}} ⟹ I = (π + 6) \int_{0}^{1} \frac{d t}{1 + (\sqrt{3} t)^{2}}

I = (π + 6) {[\frac{1}{\sqrt{3}} \arctan (\sqrt{3} t)]}_{0}^{1}

I = \frac{π + 6}{\sqrt{3}} (\arctan (\sqrt{3}) - \arctan (0)) = \frac{π + 6}{\sqrt{3}} (\frac{π}{3} - 0)

I = \frac{π (π + 6)}{3 \sqrt{3}}

💡 Teacher's Secret Hint

Be careful with the limits of integration and the constant factor from the substitution.

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Maths Question 18 – JEE-MAIN 2025

Clue 1 of 3: King's Property for Definite Integrals

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