We have,
∫dx2−3sin2x−4cos2xdx
⇒∫dx2−3sin2x−3cos2x−cos2xdx
⇒∫dx2−3(sin2x+cos2x)−cos2xdx
⇒∫dx2−3−cos2x
⇒∫dx−1−cos2x
⇒−1∫dx1+cos2x
Now, using formula,
∫1√1+x2dx=log∣∣x+√1+x2∣∣
Then,
−∫dx1+cos2x=−log∣∣cosx+√1+cos2x∣∣+C
Evaluating ∫π20cos2xdx1+3sin2x