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Question

Evaluate the integral
0e4xcos3xdx

A
325
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B
425
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C
125
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D
725
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Solution

The correct option is B 425

0e4xcos3xdx

I=e4xcos3xdx=cos3xe4xdx34+sin3xe4xdx

=cos3xe434sin3xe4xdx

e4sin3xdx=sin3x4e4x+34cos3xe4xdx

e4xcos3xdx=(cos3x)e4x434[sin3x(e4x)4+34cos3xe4x]

I+916I=e4x4[cos3x+34sin3x]

I=425[e4x](cos3x34sin3x)

0e4xcos3xdx=425e4x[cos3x+34sin3x]0

=425


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