If I(x)=∫ e^(sin²x)(cos x sin 2x-sin x)dx and I(0)=1, then I(π/3) is equal to :

Solution

Maths sir · 4 steps

Spot the pattern: the bracket is f' plus f times the derivative of sin² x, with f = cos x.

1. Look for the pattern. The derivative of sin² x is sin 2x. By the product rule, d/dx[e^(sin² x) f(x)] = e^(sin² x)(f(x)sin 2x + f'(x)).
2. Choose f. Take f(x) = cos x, so f'(x) = -sin x. Then f sin 2x + f' = cos x sin 2x - sin x, which is exactly the bracket in the integrand.

   I(x) = ∫ e^(sin² x)(cos x sin 2x - sin x)dx = e^(sin² x)cos x + C

3. Fix the constant. I(0) = e⁰cos 0 + C = 1 + C. Since I(0) = 1, we get C = 0.
4. So. I(π/3) = e^(sin²(π/3))cos π/3 = e^(3/4) × 1/2 = 1/2e^(3/4), which is option (d).

The idea to keep. When the integrand is eᵍ⁽ˣ⁾ times a bracket, test whether the bracket is f g' + f'. If it is, the answer is eᵍ⁽ˣ⁾ f(x) + C at once, with no long integration by parts.

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