The value of the definite integral ∫_(π /24)^(5π /24) dx/(1 + ∛(tan 2x)) is :

Solution

Maths sir · 4 steps

Limits add to π/4, so swap x for π/4-x and tan 2x turns into cot 2x.

1. Use King's rule. Here a + b = π/24 + 5π/24 = π/4. Replace x by π/4 - x. Then tan 2x becomes tan(π/2 - 2x) = cot 2x.
2. Rewrite with tan. With t = ∛(tan 2x), the new integrand is 1/(1 + 1/t) = t/(t + 1). So the same integral I also equals ∫_(π/24)^(5π/24) t/(1 + t) dx.
3. Add the two forms. Add 1/(1+t) and t/(1+t). The sum is exactly 1, so the hard part vanishes.

   2I = ∫_(π/24)^(5π/24) 1 dx = 5π/24 - π/24 = π/6

4. So. I = π/12, which is option (c).

The idea to keep. If the integrand is 1/(1 + f) and the swap x → a + b - x turns f into 1/f, add the two forms. You get 2I = b - a, so I = (b-a)/2.

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