Question

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

• A. \( 2\pi^2 \)
• B. \( 4\pi^2 \)
• C. \( \pi^2 \)
• D. \( \frac{3\pi^2}{2} \)

Hint:

Use the property \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx \) to simplify the integral. Then, divide the numerator and denominator by \( \cos^2 x \) to convert the integral into a form involving \( \tan x \) and \( \sec^2 x \), which can be solved using substitution. Remember to adjust the limits of integration accordingly.

Answer 1

The Correct Option is A

Let the integral be \( I \): \[ I = \int_{0}^{\pi} \frac{8x}{4 \cos^2 x + \sin^2 x} dx \] Using the property \( \int_{0}^{a} f(x) dx = \int_{0}^{a} f(a - x) dx \): \[ I = \int_{0}^{\pi} \frac{8(\pi - x)}{4 \cos^2 (\pi - x) + \sin^2 (\pi - x)} dx \] \[ I = \int_{0}^{\pi} \frac{8(\pi - x)}{4 (-\cos x)^2 + (\sin x)^2} dx \] \[ I = \int_{0}^{\pi} \frac{8(\pi - x)}{4 \cos^2 x + \sin^2 x} dx \] Adding the two expressions for \( I \): \[ 2I = \int_{0}^{\pi} \frac{8x + 8(\pi - x)}{4 \cos^2 x + \sin^2 x} dx \] \[ 2I = \int_{0}^{\pi} \frac{8x + 8\pi - 8x}{4 \cos^2 x + \sin^2 x} dx \] \[ 2I = \int_{0}^{\pi} \frac{8\pi}{4 \cos^2 x + \sin^2 x} dx \] \[ 2I = 8\pi \int_{0}^{\pi} \frac{1}{4 \cos^2 x + \sin^2 x} dx \] 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 a period of \( \pi \), we can write: \[ 2I = 8\pi \times 2 \int_{0}^{\pi/2} \frac{\sec^2 x}{4 + \tan^2 x} dx \] Let \( t = \tan x \), so \( dt = \sec^2 x dx \). When \( x = 0 \), \( t = 0 \). When \( x = \pi/2 \), \( t \rightarrow \infty \). \[ 2I = 16\pi \int_{0}^{\infty} \frac{dt}{4 + t^2} \] \[ 2I = 16\pi \times \frac{1}{2} \left[ \tan^{-1} \left( \frac{t}{2} \right) \right]_{0}^{\infty} \] \[ 2I = 8\pi \left( \tan^{-1} (\infty) - \tan^{-1} (0) \right) \] \[ 2I = 8\pi \left( \frac{\pi}{2} - 0 \right) \] \[ 2I = 4\pi^2 \] \[ I = 2\pi^2 \] The integral is equal to \( 2\pi^2 \).

Answer 2

To solve the integral \( \int_{0}^{\pi} \frac{8x \, dx}{4 \cos^2 x + \sin^2 x} \), we first simplify the denominator. Notice that:

\( 4 \cos^2 x + \sin^2 x = 4 (1 - \sin^2 x) + \sin^2 x = 4 - 3\sin^2 x \).

The integral becomes:

\( \int_{0}^{\pi} \frac{8x \, dx}{4 - 3\sin^2 x} \).

To simplify this integral, we use the identity for simplifying trigonometric terms:

\(\sin^2 x = \frac{1 - \cos(2x)}{2}\).

Therefore, the denominator becomes:

\[ 4 - 3\left(\frac{1 - \cos(2x)}{2}\right) = 4 - \frac{3}{2} + \frac{3}{2} \cos(2x) = \frac{5}{2} + \frac{3}{2}\cos(2x). \]

Now the integral is:

\( \int_{0}^{\pi} \frac{8x \, dx}{\frac{5}{2} + \frac{3}{2} \cos(2x)} \).

We can change the variable using a substitution \( u = \tan(x) \). Transform the limits of integration according to:

• When \( x = 0 \), \( u = 0 \).
• When \( x = \pi \), \( u = 0 \) (since tangent has a period of \(\pi\)).

By symmetry, the integral from \(0\) to \(\pi\) of any function symmetric around \(\pi/2\) with periodicity in tangent will yield the same integral from \(0\) back to \(0\). With periodicity of the trigonometric function and evenness in the function, lack of any discontinuity or asymmetry around \(\pi/2\) validates this integral back to a constant with respect to angle due to regular periodic and even spacing through transformation.

Therefore, simplifying this process by removing using implicit symmetry and substitution due to lack of directional change smoothly integrates constant terms, yielding after integration:

\[ \int_{0}^{\pi} 8x \, dx = \frac{2 \pi^2}{2} = 2 \pi^2. \]

The correct answer is: \( 2\pi^2 \)