Question

The value of $\int_{0}^{\pi/2}\frac{dx}{(4cos^2x+9sin^2x)^2}$ is equal to

• A. (1) $\frac{11\pi}{864}$
• B. (2) $\frac{13\pi}{864}$
• C. (3) $\frac{17\pi}{864}$
• D. (4) $\frac{26\pi}{864}$

Answer 1

Answer: (B)

(2) $\frac{13\pi}{864}$

Answer 2

We can use the formula $\int_{0}^{\pi/2}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^2} = \frac{\pi(a^2+b^2)}{4a^3b^3}$. Here, $a^2=4$ and $b^2=9$, so $a=2$ and $b=3$. Substituting these values into the formula: $\frac{\pi(4+9)}{4(2^3)(3^3)} = \frac{\pi(13)}{4(8)(27)} = \frac{13\pi}{4(216)} = \frac{13\pi}{864}$