The integral 0^ 8x dx4 ^2 x + ^2 x is equal to — Solution | JEE Mathematics

MCQMediumJEE 2025Properties of Definite Integrals

The integral $\int_{0}^{\pi} \frac{8x \, dx}{4 \cos^2 x + \sin^2 x}$ is equal to

Answer

Correct answer:A

Standard Method

Given:

I = \int_{0}^{\pi} \frac{8x}{4 \cos^2 x + \sin^2 x} \, dx

Find: The value of the integral and hence the correct option.

Use the property

\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx

So,

I = \int_{0}^{\pi} \frac{8(\pi - x)}{4 \cos^2 (\pi - x) + \sin^2 (\pi - x)} \, dx

Since $\cos(\pi-x) = -\cos x$ and $\sin(\pi-x) = \sin x$ ,

I = \int_{0}^{\pi} \frac{8(\pi - x)}{4 \cos^2 x + \sin^2 x} \, dx

Add the two forms and substitute

Adding the two expressions for $I$ ,

2I = \int_{0}^{\pi} \frac{8x + 8(\pi-x)}{4 \cos^2 x + \sin^2 x} \, dx

2I = \int_{0}^{\pi} \frac{8\pi}{4 \cos^2 x + \sin^2 x} \, dx

2I = 8\pi \int_{0}^{\pi} \frac{1}{4 \cos^2 x + \sin^2 x} \, dx

Divide numerator and denominator by $\cos^2 x$ :

2I = 8\pi \int_{0}^{\pi} \frac{\sec^2 x}{4 + \tan^2 x} \, dx

Since the integrand has period $\pi$ ,

2I = 16\pi \int_{0}^{\pi/2} \frac{\sec^2 x}{4 + \tan^2 x} \, dx

Let $t = \tan x$ , so $dt = \sec^2 x \, dx$ . Then when $x=0$ , $t=0$ , and when $x \to \pi/2$ , $t \to \infty$ .

2I = 16\pi \int_{0}^{\infty} \frac{dt}{4+t^2}

2I = 16\pi \times \frac{1}{2} \left[ \tan^{-1}\left(\frac{t}{2}\right) \right]_{0}^{\infty}

2I = 8\pi \left( \frac{\pi}{2} - 0 \right) = 4\pi^2

Hence,

I = 2\pi^2

Therefore, the integral is $2\pi^2$ and the correct option is A.

Using the symmetry property incorrectly. The relation $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$ must be applied to the entire integrand. Do not replace only $x$ by $\pi-x$ and ignore the trigonometric terms.
Dividing by $\cos^2 x$ carelessly. After division, $\frac{1}{4\cos^2 x + \sin^2 x}$ becomes $\frac{\sec^2 x}{4+\tan^2 x}$ , not $\frac{1}{4+\tan^2 x}$ . The factor $\sec^2 x$ is essential for substitution.
Making an incorrect substitution limit at $x=\pi$ directly for $t=\tan x$ . The working uses periodicity to reduce the interval to $[0,\pi/2]$ first. Substituting over $[0,\pi]$ without handling the discontinuity at $\pi/2$ leads to errors.

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