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Question

Find the integral of the function
sin3x+cos3xsin2xcos2x

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Solution

Consider the given integral.


I=sin3x+cos3xsin2xcos2xdx


I=sin3xsin2xcos2xdx+cos3xsin2xcos2xdx


I=sinxcos2xdx+cosxsin2xdx


I=I1+I2 ……. (1)



Let I1=sinxcos2xdx



Put t=cosx


dtdxsinx


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=1cosx1sinx+C



Hence, this is the answer.


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