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Evaluate: int(cos2x-cos2alpha)/(cosx-cos...

Evaluate: `int(cos2x-cos2alpha)/(cosx-cosalpha)\ dx`

Text Solution

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The correct Answer is:
`sinx+x"cos"theta+C`
`int(cos2x-cos2theta)/(cosx-costheta)dx=int(cos^(2)x-cos^(2)theta)/(cosx-"cos"theta)dx`
`=int(cosx+"cos"theta)dx`
`=sinx+x"cos"theta+C`
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