The value of ∫cos3⁡x+cos5⁡xsin2⁡x+sin4⁡xdx is

Since the integrands satisfies. R(sin⁡x,−cos∣x)=−R(sin⁡x,cos⁡x)Therefore, we substitute sin⁡x=t, so that cos⁡xdx=dt.∫cos3⁡x+cos5⁡xsin2⁡x+sin4⁡xdx=∫cos⁡x1−sin2⁡x+1−sin2⁡x2sin2⁡x+sin4⁡xdx=∫1−t22−t2t2+t4dt=∫1+2t2−61+t2dt=t−2t−6tan−1⁡t+C=sin⁡x−2(sin⁡x)−1−6tan−1⁡(sin⁡x)+C.