We can start by using integration by parts, with:
u = sin(2x) dv = tan⁻¹(sin x) dx
Then:
du/dx = 2cos(2x) v = x/2 tan⁻¹(sin x) – (1/4) ln(cos x)
Using these substitutions, we have:
x/2 f0 sin 2x tan-1(sin x) dx = [x/2 sin(2x) [x/2 tan⁻¹(sin x) – (1/4) ln(cos x)]]₀ᴨ/2 – f0 [x/2 tan⁻¹(sin x) – (1/4) ln(cos x)] cos(2x) dx
Next, we use integration by parts again, with:
u = x/2 tan⁻¹(sin x) – (1/4) ln(cos x) dv = cos(2x) dx
Then:
du/dx = (1/2) [tan⁻¹(sin x) – x sin x/(cos x)²] + (1/4) sin x/cos x v = (1/2) sin(2x)
Substituting these values, we get:
x/2 f0 sin 2x tan-1(sin x) dx = [x/2 sin(2x) [x/2 tan⁻¹(sin x) – (1/4) ln(cos x)]]₀ᴨ/2 – f0 [(1/2) sin(2x)] [x/2 tan⁻¹(sin x) – (1/4) ln(cos x)]₀ᴨ/2 – f0 ᴨ/2 [1/2 tan⁻¹(sin ᴨ/2) – (1/4) ln(cos ᴨ/2)]
Simplifying, we get:
x/2 f0 sin 2x tan-1(sin x) dx = (1/4) ᴨ ln 2 – f0 [(1/2) sin(2x) tan⁻¹(sin x) – (1/8) ln(cos²x)]₀ᴨ/2 – (1/4) ᴨ tan⁻¹(1)
Therefore, the answer is (1/4) ᴨ ln 2 – (1/2) f0 [sin(2x) tan⁻¹(sin x) – (1/4) ln(cos²x)]₀ᴨ/2 – (1/4) ᴨ tan⁻¹(1).
