The integral∫ (1-1/√3)(cos x-sin x)/(1+2/√3 sin 2x)dx is equal to

Solution

Maths sir · 4 steps

Put t = sin x + cos x: the numerator becomes exactly dt and sin 2x becomes t² - 1.

1. Substitute. Let t = sin x + cos x, so dt = (cos x - sin x) dx. Squaring gives t² = 1 + sin 2x, so sin 2x = t² - 1.
2. Tidy the denominator. 1 + 2/√3(t² - 1) = 2/√3(t² - a²) with a² = 1 - √3/2, which gives a = (√3-1)/2. The constant in front becomes (1 - 1/√3)√3/2 = (√3-1)/2, which is exactly a.

   ∫ (a dt)/(t² - a²) = a × 1/2a ln|(t-a)/(t+a)| + C = 1/2 ln|(t-a)/(t+a)| + C

3. Back to x. Write t = √2 sin(x + π/4) and a = √2 sin π/12. Sum-to-product gives t - a = 2√2 cos(x/2+π/6)sin(x/2+π/12) and t + a = 2√2 sin(x/2+π/6)cos(x/2+π/12).
4. So. (t-a)/(t+a) = tan(x/2+π/12)/tan(x/2+π/6), and the integral is 1/2 logₑ of the absolute value of this ratio, plus C. That is option (a).

The idea to keep. Numerator cos x - sin x with sin 2x below means put t = sin x + cos x. Then write the constant a as √2 sin θ so sum-to-product turns the log into tangents.

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