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