Question

Evaluate `int_0^(pi)e^2x.sin(pi/4+x)dx`

Answer 1

Let `I==int_0^pie^(2x)sin(pi/2+x)dx`

Integrating by parts, we get

` I=1/2[e^(2x)sin(pi/4+x)_0^pi]-1/2int_0^pie^(2x)cos(pi/4+x)dx`

 Now, integrating the second term by parts, we get

` =>I=1/2[e^(2x)sin(pi/4+x)_0^pi]-1/2{[1/2e^(2x)cos(pi/4+x)_0^pi]+1/2int_0^pi e^(2x)sin(pi/4+x)dx}`

=>`I=1/2[e^(2x)sin(pi/4+x)_0^pi]-1/4[e^(2x)cos(pi/4+x)_0^pi]-1/4I`

`=>5/4I=1/2[e^(2x)sin(pi+pi/4)-sin(pi/4)]-1/4[e^(2x)cos(pi+pi/4)-cos(pi/4)]`

`=>5/4I=1/2 |__-e^(2x)xx1/sqrt2-1/sqrt2__|-1/4|__-e^(2pi)xx1/sqrt2-1/sqrt2__|`

`=>5/4I==1/(2sqrt2)e^(2pi)-1/(2sqrt2)+1/(4sqrt2)e^(2pi)+1/(4sqrt2)`

`=>I=-1/(5sqrt2)(e^(2pi)+1)`