Consider the given integral.
I=∫sin3x+cos3xsin2xcos2xdx
I=∫sin3xsin2xcos2xdx+∫cos3xsin2xcos2xdx
I=∫sinxcos2xdx+∫cosxsin2xdx
I=I1+I2 ……. (1)
Let I1=∫sinxcos2xdx
Put t=cosx
dtdx−sinx
−dt=sinxdx
Therefore,
I1=−∫1t2dt
I1=1t+C
On putting the value of t, we get
I1=1cosx+C
Now,
I2=∫cosxsin2xdx
Let u=sinx
du=cosxdx
Therefore,
I2=∫1u2du
I2=−1u+C
Put the value of u, we get
I2=−1sinx+C
From equation (1),
I=1cosx−1sinx+C
Hence, this is the answer.