Question

$\int_0^{\pi/2} \frac{1}{5\cos^2 x + 16\sin^2 x + 8\sin x \cos x} dx =$

• A. $\text{Tan}^{-1}\left(\frac{4}{5}\right)$
• B. $2\text{Tan}^{-1}\left(\frac{3}{5}\right)$
• C. $\frac{1}{8}\text{Tan}^{-1}\left(\frac{8}{9}\right)$
• D. $\frac{1}{4}\text{Tan}^{-1}\left(\frac{7}{8}\right)$
• E. None of these

Hint:

This type of integral, $\int \frac{dx}{a\cos^2 x + b\sin^2 x + c\sin x \cos x}$, is a standard form. The method of dividing by $\cos^2 x$ and substituting $u=\tan x$ is almost always the most efficient way to solve it. Be careful when completing the square and evaluating the final arctan integral.

Answer 1

Answer: (E)

Step 1: Understanding the Concept
This is a definite integral of a trigonometric function. The integrand is of a standard form that can be solved by dividing the numerator and denominator by $\cos^2 x$ and then using the substitution $u = \tan x$.
Step 2: Key Formula or Approach
1. Divide the numerator and denominator of the integrand by $\cos^2 x$. The numerator becomes $\sec^2 x$. 2. The integral transforms into the form $\int \frac{\sec^2 x}{A\tan^2 x + B\tan x + C} dx$. 3. Use the substitution $u = \tan x$, so $du = \sec^2 x dx$. 4. Change the limits of integration according to the substitution. 5. Evaluate the resulting integral of a rational function, which will likely be of the form $\int \frac{du}{au^2+bu+c}$, by completing the square.
Step 3: Detailed Explanation
Let $I = \int_0^{\pi/2} \frac{1}{5\cos^2 x + 16\sin^2 x + 8\sin x \cos x} dx$. 1. Transform the integrand: Divide numerator and denominator by $\cos^2 x$: \[ I = \int_0^{\pi/2} \frac{\sec^2 x}{5 + 16\tan^2 x + 8\tan x} dx \] 2. Substitution: Let $u = \tan x$. Then $du = \sec^2 x dx$. Change the limits: When $x=0$, $u = \tan(0) = 0$. When $x \to \pi/2$, $u = \tan(x) \to \infty$. The integral becomes: \[ I = \int_0^\infty \frac{1}{16u^2 + 8u + 5} du \] 3. Complete the square: The denominator is $16u^2 + 8u + 5$. \[ 16u^2 + 8u + 5 = 16\left(u^2 + \frac{1}{2}u\right) + 5 = 16\left(u^2 + \frac{1}{2}u + \frac{1}{16} - \frac{1}{16}\right) + 5 \] \[ = 16\left(u+\frac{1}{4}\right)^2 - 1 + 5 = 16\left(u+\frac{1}{4}\right)^2 + 4 \] 4. Evaluate the integral: \[ I = \int_0^\infty \frac{1}{16(u+1/4)^2 + 4} du \] Let $v = u+1/4$, so $dv=du$. The limits change from $[0, \infty)$ for $u$ to $[1/4, \infty)$ for $v$. \[ I = \int_{1/4}^\infty \frac{1}{16v^2 + 4} dv = \frac{1}{4} \int_{1/4}^\infty \frac{1}{4v^2+1} dv = \frac{1}{4} \int_{1/4}^\infty \frac{1}{(2v)^2+1} dv \] This is a standard arctangent integral form. \[ I = \frac{1}{4} \left[ \frac{1}{2} \arctan(2v) \right]_{1/4}^\infty = \frac{1}{8} [\arctan(2v)]_{1/4}^\infty \] \[ I = \frac{1}{8} \left( \lim_{v\to\infty} \arctan(2v) - \arctan\left(2 \cdot \frac{1}{4}\right) \right) \] \[ I = \frac{1}{8} \left( \frac{\pi}{2} - \arctan\left(\frac{1}{2}\right) \right) \] Using the identity $\arctan(x) + \text{arccot}(x) = \pi/2$, we get $\pi/2 - \arctan(1/2) = \text{arccot}(1/2) = \arctan(2)$. So, the final result is: \[ I = \frac{1}{8}\arctan(2) \] This result does not match any of the options provided. Step 4: Final Answer
The correct value of the integral is $\frac{1}{8}\arctan(2)$. As this is not among the options, the question is likely flawed.