If ∫ 1/(a²sin² x+b²cos² x) dx=1/12 tan⁻¹(3 tan x)+ constant, then the maximum value of a sin x+b cos x, is :

Solution

Maths sir · 4 steps

Divide top and bottom by cos² x, put t = tan x, and match the result with the given answer.

1. Convert to tan. Divide the numerator and the denominator by cos² x. The integrand becomes (sec² x)/(a² tan² x + b²), and sec² x dx is exactly dt for t = tan x.

   ∫ dt/(a² t² + b²) = 1/ab tan⁻¹((a t)/b) + C

2. Compare. Matching with 1/12 tan⁻¹(3 tan x) gives ab = 12 and a/b = 3. Putting a = 3b into ab = 12 gives 3b² = 12, so b² = 4 and a² = 9b² = 36.
3. Maximum value. The maximum of a sin x + b cos x is √(a² + b²) = √(36 + 4).
4. So. The maximum value is √40, option (c).

The idea to keep. When you see a²sin² x + b²cos² x in a denominator, divide by cos² x and put t = tan x. And always remember: a sin x + b cos x peaks at √(a² + b²).

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