Let for f(x)=7 tan ⁸ x+7 tan ⁶ x-3 tan ⁴ x-3 tan ² x, I₁=∫₀^(π / 4) f(x) d x and I₂=∫₀^(π / 4) x f(x) d x. Then 7 I₁+12 I₂ is equal to :

Solution

Maths sir · 5 steps

Pull out sec² x from f(x) and it turns into a perfect derivative, so both integrals collapse.

1. Factor f(x). Group the terms in pairs: 7 tan⁶ x(tan² x+1) - 3 tan² x(tan² x+1). Since 1+tan² x = sec² x, this is f(x) = (7 tan⁶ x - 3 tan² x)sec² x.

   f(x) = d/dx(tan⁷ x - tan³ x)

2. Find I1. Call F(x) = tan⁷ x - tan³ x. Then I₁ = F(π/4) - F(0) = (1-1) - 0 = 0.
3. Find I2 by parts. I₂ = [xF(x)]₀^(π/4) - ∫₀^(π/4) F(x) dx. The boundary term is 0 because F(π/4) = 0.
4. Integrate F. Write tan⁷ x - tan³ x = (tan⁵ x - tan³ x)sec² x. Put t = tan x: ∫₀¹ (t⁵ - t³) dt = 1/6 - 1/4 = -1/12, so I₂ = 1/12.
5. So. 7I₁ + 12I₂ = 7(0) + 12 × 1/12 = 1, which is option (b).

The idea to keep. When even powers of tan x come in neighbouring pairs, factor out 1+tan² x = sec² x. The integrand then becomes the derivative of a polynomial in tan x.

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