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Find the following integral without using BC methods like integration by parts.  ...

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thisuserhasma... | Student, Grade 10 | (Level 1) Valedictorian

Posted May 7, 2013 at 6:50 PM via web

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Find the following integral without using BC methods like integration by parts.

 

`int_-2^2 4/(x^2+4)dx`

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted May 7, 2013 at 11:50 PM (Answer #2)

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The integral `int_(-2)^2 4/(x^2+4) dx` can be determined by using the formula `int 1/(x^2 + a^2) = (tan^-1(x/a))/a`

`int_(-2)^2 4/(x^2+4) dx`

= `[4*(tan^-1(x/2))/2]_-2^2`

= `[2*tan^-1(x/2)]_-2^2`

= `2*tan^-1(2/2) - 2*tan^-1(-2/2)`

= `pi/2 + pi/2`

= `pi`

The definite integral `int_(-2)^2 4/(x^2+4) dx = pi`

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oldnick | (Level 1) Valedictorian

Posted May 8, 2013 at 2:51 AM (Answer #3)

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The integral can be determined by using the formula

=   WROOOOOOOOOOONG!

`d/dx 2tan^(-1) (x/2)=2 1/(1+(x/2)^2)=` `2 1/(1+x^2/4)=` `2 1/((4+x^2)/4)=` `8/(4+x^2)`   WROOOOOOOOOOOOOOOOONG!

`int_(-2)^2 4/(4+x^2) dx= pi/2` ! no `pi!`

 

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oldnick | (Level 1) Valedictorian

Posted May 7, 2013 at 11:32 PM (Answer #1)

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`int_(-2)^2 4/(x^2+4) dx` `=int_(-2)^2 1/((x/2)^2+1) dx` `= [arc tan(x/2)]_(-2)^2=` `=3/4 pi- pi/4=pi/2`

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