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Question

solve: 1cos(xa)cos(xb) dx

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Solution

We have,

1cos(xa)cos(xb)dx


Divide and multiply by sin(ab)

Then,

sin(ab)sin(ab)×1cos(xa)cos(xb)dx

=1sin(ab)sin(ab)cos(xa)cos(xb)dx

=1sin(ab)sin(ab+xx)cos(xa)cos(xb)dx

=1sin(ab)sin(xb+ax)cos(xa)cos(xb)dx

=1sin(ab)sin[(xb)+(ax)]cos(xa)cos(xb)dx


Using formula sin(AB)=sinAcosBcosAsinB

So,

1sin(ab)sin(xb)cos(xa)cos(xb)sin(xa)cos(xa)cos(xb)dx

=1sin(ab)[sin(xb)cos(xa)cos(xa)cos(xb)dxcos(xb)sin(xa)cos(xa)cos(xb)dx]

=1sin(ab)[sin(xb)cos(xb)dxsin(xa)cos(xa)dx]

=1sin(ab)[tan(xb)dxtan(xa)dx]

=1sin(ab)[log|cos(xb)|+log|cos(xa)|]

=1sin(ab)[log|cos(xa)|log|cos(xb)|]

=1sin(ab)[log|cos(xa)||cos(xb)|]


Hence, this is the answer.


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