I1=∫sin2(2x−1)dx
$=\dfrac{1}{2}\int [1-\cos (4x-2) ] dx = \dfrac{1}{2}x -\dfrac{1}{2.4}\int\cos (4x-2) d(4x-2)=\dfrac{1}{2}x-\dfrac{1}{8}\sin (4x-2)+C$
I2=∫dx(cosx−3√sinx)2
$=\int\dfrac{dx}{4(\dfrac{1}{2}\cos x -\dfrac{\sqrt 3}{2}\sin x)^2} =\dfrac{1}{4}\int \dfrac{dx}{\cos^2 (x+\dfrac{\pi}{3})}$
$=\dfrac{1}{4}\int \dfrac{d(x+\dfrac{\pi}{3})}{\cos^2 (x+\dfrac{\pi}{3})}=\dfrac{1}{4}\tan (x+\dfrac{\pi}{3})+C$