Question

If $\int_0^{\frac{\pi}{2}} \frac{dx}{5+4\sin x} = A \tan^{-1} B$, then A + B =

• A. $\frac{2}{3}$
• B. 1
• C. 2
• D. $\frac{1}{3}$

Hint:

When dealing with Weierstrass substitution conversions, always take extra care when evaluating the fraction arithmetic at the boundaries. Keeping fractional terms organized on your scratchpad prevents simple arithmetic slips!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a definite integral identity containing a trigonometric denominator. We need to solve the integral to find the values of constants $A$ and $B$, and then compute their sum.

Step 2: Key Formula or Approach:
We can evaluate this standard trigonometric integral template by using Weierstrass half-angle substitutions: $$ \text{Put } \tan\left(\frac{x}{2}\right) = t \implies dx = \frac{2dt}{1+t^2} \quad \text{and} \quad \sin x = \frac{2t}{1+t^2} $$

Step 3: Detailed Explanation:
Let's first adjust the integration limits based on our half-angle substitution variable $t = \tan\left(\frac{x}{2}\right)$:

• When $x = 0 \implies t = \tan(0) = 0$

• When $x = \frac{\pi}{2} \implies t = \tan\left(\frac{\pi}{4}\right) = 1$
Substituting these expressions back into our integral $I$: $$ I = \int_0^1 \frac{\frac{2dt}{1+t^2}}{5 + 4\left(\frac{2t}{1+t^2}\right)} $$ Multiply through by the common denominator $(1+t^2)$ to simplify: $$ I = \int_0^1 \frac{2dt}{5(1+t^2) + 8t} = \int_0^1 \frac{2dt}{5t^2 + 8t + 5} $$ Factor out 5 from the denominator to prepare for completing the square: $$ I = \frac{2}{5} \int_0^1 \frac{dt}{t^2 + \frac{8}{5}t + 1} $$ Complete the square for the quadratic expression in the denominator by adding and subtracting $\left(\frac{4}{5}\right)^2 = \frac{16}{25}$: $$ t^2 + \frac{8}{5}t + 1 = \left(t + \frac{4}{5}\right)^2 + 1 - \frac{16}{25} = \left(t + \frac{4}{5}\right)^2 + \frac{9}{25} = \left(t + \frac{4}{5}\right)^2 + \left(\frac{3}{5}\right)^2 $$ Substitute this completed square back into our integral: $$ I = \frac{2}{5} \int_0^1 \frac{dt}{\left(t + \frac{4}{5}\right)^2 + \left(\frac{3}{5}\right)^2} $$ Apply the standard integral formula $\int \frac{dt}{t^2+a^2} = \frac{1}{a}\tan^{-1}\left(\frac{t}{a}\right)$: $$ I = \frac{2}{5} \cdot \frac{1}{\frac{3}{5}} \left[ \tan^{-1}\left( \frac{t + \frac{4}{5}}{\frac{3}{5}} \right) \right]_0^1 = \frac{2}{3} \left[ \tan^{-1}\left( \frac{5t + 4}{3} \right) \right]_0^1 $$ Now, substitute the upper limit 1 and lower limit 0: $$ I = \frac{2}{3} \left[ \tan^{-1}\left(\frac{5(1)+4}{3}\right) - \tan^{-1}\left(\frac{5(0)+4}{3}\right) \right] = \frac{2}{3} \left[ \tan^{-1}(3) - \tan^{-1}\left(\frac{4}{3}\right) \right] $$ Apply the subtraction identity for inverse tangents $\tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$: $$ I = \frac{2}{3} \tan^{-1}\left[ \frac{3 - \frac{4}{3}}{1 + 3\left(\frac{4}{3}\right)} \right] = \frac{2}{3} \tan^{-1}\left[ \frac{\frac{5}{3}}{1 + 4} \right] = \frac{2}{3} \tan^{-1}\left[ \frac{5}{3 \times 5} \right] = \frac{2}{3} \tan^{-1}\left(\frac{1}{3}\right) $$ Comparing this result with the given statement $A \tan^{-1} B$: $$ A = \frac{2}{3} \quad \text{and} \quad B = \frac{1}{3} $$ Calculating their sum: $$ A + B = \frac{2}{3} + \frac{1}{3} = \frac{3}{3} = 1 $$

Step 4: Final Answer:
The value of $A + B$ is 1, which corresponds to option (B).