Question

The integral $\int_0^\pi \frac{8xdx}{4\cos^2x + \sin^2x}$ is equal to

• A. $2\pi^2$
• B. $4\pi^2$
• C. $\pi^2$

Answer 1

Answer: (A)

2\pi^2

Answer 2

The integral to be evaluated is $I = \int_0^\pi \frac{8xdx}{4\cos^2x + \sin^2x}$.

Step 1: Apply the property of definite integrals $\int_0^a f(x) dx = \int_0^a f(a-x) dx$.
Let the given integral be $I$. Here, $a = \pi$. $I = \int_0^\pi \frac{8(\pi-x)dx}{4\cos^2(\pi-x) + \sin^2(\pi-x)}$
Since $\cos(\pi-x) = -\cos x$ and $\sin(\pi-x) = \sin x$, we have $\cos^2(\pi-x) = \cos^2 x$ and $\sin^2(\pi-x) = \sin^2 x$.
So, $I = \int_0^\pi \frac{8(\pi-x)dx}{4\cos^2x + \sin^2x}$

Step 2: Add the original integral and the transformed integral.
Adding the original integral $I = \int_0^\pi \frac{8xdx}{4\cos^2x + \sin^2x}$ to the transformed integral:
$2I = \int_0^\pi \frac{8xdx}{4\cos^2x + \sin^2x} + \int_0^\pi \frac{8(\pi-x)dx}{4\cos^2x + \sin^2x}$
$2I = \int_0^\pi \frac{8x + 8(\pi-x)}{4\cos^2x + \sin^2x} dx$
$2I = \int_0^\pi \frac{8x + 8\pi - 8x}{4\cos^2x + \sin^2x} dx$
$2I = \int_0^\pi \frac{8\pi}{4\cos^2x + \sin^2x} dx$
$2I = 8\pi \int_0^\pi \frac{dx}{4\cos^2x + \sin^2x}$
$I = 4\pi \int_0^\pi \frac{dx}{4\cos^2x + \sin^2x}$

Step 3: Evaluate the new integral $J = \int_0^\pi \frac{dx}{4\cos^2x + \sin^2x}$.
Let $f(x) = \frac{1}{4\cos^2x + \sin^2x}$. We observe that $f(\pi-x) = \frac{1}{4\cos^2(\pi-x) + \sin^2(\pi-x)} = \frac{1}{4\cos^2x + \sin^2x} = f(x)$.
This allows us to use the property $\int_0^{2a} f(x) dx = 2\int_0^a f(x) dx$ where $2a=\pi$, so $a=\pi/2$.
$J = 2 \int_0^{\pi/2} \frac{dx}{4\cos^2x + \sin^2x}$

Step 4: Simplify the integrand by dividing the numerator and denominator by $\cos^2x$.
$J = 2 \int_0^{\pi/2} \frac{\frac{1}{\cos^2x}}{\frac{4\cos^2x}{\cos^2x} + \frac{\sin^2x}{\cos^2x}} dx$
$J = 2 \int_0^{\pi/2} \frac{\sec^2x}{4 + \tan^2x} dx$

Step 5: Use substitution to evaluate the integral.
Let $t = \tan x$. Then $dt = \sec^2x dx$.
When $x=0$, $t = \tan 0 = 0$.
When $x=\pi/2$, $t = \tan(\pi/2) \to \infty$.
$J = 2 \int_0^\infty \frac{dt}{4 + t^2}$
This is a standard integral of the form $\int \frac{dx}{a^2+x^2} = \frac{1}{a}\arctan\left(\frac{x}{a}\right)$. Here $a^2=4$, so $a=2$.
$J = 2 \left[ \frac{1}{2} \arctan\left(\frac{t}{2}\right) \right]_0^\infty$
$J = \left[ \arctan\left(\frac{t}{2}\right) \right]_0^\infty$
$J = \lim_{t \to \infty} \arctan\left(\frac{t}{2}\right) - \arctan\left(\frac{0}{2}\right)$
$J = \frac{\pi}{2} - 0$
$J = \frac{\pi}{2}$

Step 6: Substitute the value of $J$ back into the expression for $I$.
$I = 4\pi J$
$I = 4\pi \left(\frac{\pi}{2}\right)$
$I = 2\pi^2$

The final answer is $2\pi^2$.