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Using integration by parts, find `inte^(3x)cos4x dx` .

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roshan-rox | (Level 1) Valedictorian

Posted August 2, 2013 at 4:06 AM via web

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Using integration by parts, find `inte^(3x)cos4x dx` .

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted August 2, 2013 at 4:40 AM (Answer #1)

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`I = inte^(3x)cos4xdx`

Let;

`U = e^(3x)`

`dU = 3e^(3x)dx`

`V = 1/4sin4x`

`dV = cos4xdx`

From integral by parts;

`intUdV = UV-intVdU`

`inte^(3x)cos4xdx = (e^(3x)xx1/4sin4x)-int1/4sin4x3e^(3x)dx`

`J = inte^(3x)sin4xdx`

Using integral by parts same as above we can get that;

`inte^(3x)sin4xdx = (e^(3x)xx(-1/4cos4x))-int(-1/4cos4x)3e^(3x)dx`

`J = (e^(3x)xx(-1/4cos4x))+3/4inte^(3x)cos4xdx`

`J = (e^(3x)xx(-1/4cos4x))+3/4I`

`inte^(3x)cos4xdx = (e^(3x)xx1/4sin4x)-int1/4sin4x3e^(3x)dx`

`I = (e^(3x)xx1/4sin4x)-3/4J`

`I = (e^(3x)xx1/4sin4x)-3/4((e^(3x)xx(-1/4cos4x))+3/4I)`

`I = e^(3x)sin(4x)/4+3/16e^(3x)cos4x-9/16I`

`I(1+9/16) = e^(3x)sin(4x)/4+3/16e^(3x)cos4x`

`25/16I = (4e^(3x)sin4x+3e^(3x)cos4x)/16`

`I = (4e^(3x)sin4x+3e^(3x)cos4x)/25`

`I = (e^(3x)(4sin4x+3cos4x))/25+C` where C is a constant.

So the answer is;

`I = (e^(3x)(4sin4x+3cos4x))/25+C `

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