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Question

Suppose 17cos2xsin7xcos2xdx=g(x)sin7x+C, where C is arbitrary constant of integration. Then the value of g(0)+g′′(π4) is

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Solution

17cos2xsin7xcos2xdx
=sec2xsin7xdx71sin7xdx=I1I2

Now, I1=(1sin7x)sec2x dx
=tanxsin7x+7tanxsin8xcosx dx=tanxsin7x+ I2
I1I2=tanxsin7x+C, where C is constant of integration.

Hence, g(x)=tanx
g(x)=sec2x and g′′(x)=2sec2xtanx
g(0)=1 and g′′(π4)=4
Hence, g(0)+g′′(π4)=5

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