Definite Integration
Integration
Grade Class 12
Question:
Let I(x) = ∫ \frac{x^2(\sec^2 x + \tan x)}{(x \tan x + 1)^2} dx. If I(0) = 0 then I\left(\frac{\pi}{4}\right) is equal to
\log_e \frac{(\pi+4)^2}{16} - \frac{\pi^2}{4(\pi+4)}
\log_e \frac{(\pi+4)^2}{16} + \frac{\pi^2}{4(\pi+4)}
\log_e \frac{(\pi+4)^2}{32} - \frac{\pi^2}{4(\pi+4)}
\log_e \frac{(\pi+4)^2}{32} + \frac{\pi^2}{4(\pi+4)}
Step-by-Step Solution
Key Concept: The integral can be solved by recognizing the derivative of a quotient or by using integration by parts. Specifically, notice that d/dx (x / (x tan x + 1)) = ( (tan x + x sec^2 x)(x tan x + 1) - x(tan x + x sec^2 x) ) / (x tan x + 1)^2 = (tan x + x sec^2 x) / (x tan x + 1)^2. This suggests rewriting the integrand as x * [x(sec^2 x + tan x) / (x tan x + 1)^2] and using integration by parts.
Let u = x and dv = (x sec^2 x + x tan x) / (x tan x + 1)^2 dx. Note that d/dx (x / (x tan x + 1)) = ( (tan x + x sec^2 x)(x tan x + 1) - x(tan x + x sec^2 x) ) / (x tan x + 1)^2 = (tan x + x sec^2 x) / (x tan x + 1)^2. This is not quite the integrand. Let's rewrite the integrand as (x^2 sec^2 x + x^2 tan x) / (x tan x + 1)^2. Let f(x) = x / (x tan x + 1). Then f'(x) = (tan x + x sec^2 x) / (x tan x + 1)^2. The integral is \int x * (x sec^2 x + x tan x) / (x tan x + 1)^2 dx. This can be solved by parts.
Correct Answer: 3