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: Divide by \( \cos^2 x \) Divide numerator and denominator by \( \cos^2 x \): \[ I = \int_0^{\pi/4} \frac{\sec^2 x}{5 + 16\tan^2 x + 8\tan x} dx \]
Step 2: Substitution Let \( t = \tan x \). \( dt = \sec^2 x dx \). Limits: \( x=0 \to t=0 \); \( x=\pi/4 \to t=1 \). \[ I = \int_0^1 \frac{dt}{16t^2 + 8t + 5} \]
Step 3: Complete the Square Denominator: \( 16(t^2 + \frac{1}{2}t + \frac{5}{16}) = 16( (t+1/4)^2 + \frac{5}{16} - \frac{1}{16} ) = 16( (t+1/4)^2 + (1/2)^2 ) \). \[ I = \frac{1}{16} \int_0^1 \frac{dt}{(t+1/4)^2 + (1/2)^2} \] Using formula \( \int \frac{dx}{x^2+a^2} = \frac{1}{a} \tan^{-1}(x/a) \): \[ I = \frac{1}{16} \cdot \frac{1}{1/2} \left[ \tan^{-1}\left(\frac{t+1/4}{1/2}\right) \right]_0^1 \] \[ I = \frac{1}{8} \left[ \tan^{-1}(2t + 0.5) \right]_0^1 \] Wait, \( \frac{t+0.25}{0.5} = 2t + 0.5 \). Upper limit \( t=1 \): \( \tan^{-1}(2.5) = \tan^{-1}(5/2) \). Lower limit \( t=0 \): \( \tan^{-1}(0.5) = \tan^{-1}(1/2) \).
Step 4: Simplify \( I = \frac{1}{8} [ \tan^{-1}(5/2) - \tan^{-1}(1/2) ] \). Using \( \tan^{-1}x - \tan^{-1}y = \tan^{-1}\frac{x-y}{1+xy} \): \( \frac{5/2 - 1/2}{1 + (5/2)(1/2)} = \frac{2}{1 + 5/4} = \frac{2}{9/4} = \frac{8}{9} \). \[ I = \frac{1}{8} \tan^{-1}\left(\frac{8}{9}\right) \]