∫2 sin x(3+sin 2x)dx is equal to
14 ln ∣∣2+sin x−cos x2−sin x+cos x∣∣−1√2tan−1(sin x+cos x√2)+C
∫2 sin x(3+sin 2x)dx
=∫sin x+cos x+sin x−cos x(3+sin 2x)dx
=∫sin x+cos x3+sin 2xdx−∫−sin x+cos x(3+sin 2x)
= I1+I2
Putting t1=sin x−cos x in I1 and t2=sin x+cos x in I2, we get
I=∫dt1(3+(1−t21))−∫dt2(3+(t22−1))=∫dt14−t21−∫dt22+t22
=14 ln ∣∣2+t12−t1∣∣−1√2tan−1(t2√2)+C
=12 ln ∣∣2+sin x−cos x2−sin x+cos x∣∣−1√2tan−1(sin x+cos x√2)+C