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Question

Integrate sin3x+cos3xsin2xcos2x

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Solution

We have,

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

I=sec x cosec x+C

Hence, this is the answer.


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