# Find the following integral without using BC methods like integration by parts. `int_-2^2 4/(x^2+4)dx`

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### 3 Answers

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` **

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!`

`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`