Let I(x)=∫ x²(x sec² x+tan x)/(x tan x+1)²dx. If I(0)=0, then I(π/4) is equal to :

Solution

Maths sir · 5 steps

The bracket on top is the derivative of x tan x+1, so take x² as the part to differentiate.

1. Spot the derivative. Let f(x) = x tan x + 1. Then f'(x) = tan x + x sec² x, which is exactly the bracket in the numerator. So the integrand is x² · f'(x)/f(x)², and ∫ f'/f² dx = -1/f.

   I(x) = -x²/(x tan x + 1) + ∫ 2x/(x tan x + 1) dx

2. Second integral. Multiply top and bottom by cos x to get (2x cos x)/(x sin x + cos x). The derivative of x sin x + cos x is sin x + x cos x - sin x = x cos x, so this integral is 2 ln|x sin x + cos x|.
3. Fix the constant. I(x) = -x²/(x tan x + 1) + 2 ln|x sin x + cos x| + C. At x = 0 both terms are 0, because ln 1 = 0. Since I(0) = 0, this gives C = 0.
4. Put in the value. At x = π/4, x tan x + 1 = (π + 4)/4, so the first term is -π²/16 × 4/(π + 4) = -π²/4(π + 4). Also x sin x + cos x = 1/√2(π/4 + 1) = (π + 4)/4√2, so 2 ln of this is ln(π + 4)²/32.
5. So. I(π/4) = ln(π + 4)²/32 - π²/4(π + 4), which is option (a).

The idea to keep. When a numerator holds the derivative of the squared denominator, ∫ f'/f² dx = -1/f is waiting. Pair it with integration by parts on the leftover factor.

Answer from the NTA final answer key. Solution by PrepWiser.