`int sec^2 (x/2)tan(x/2) dx` Find the indefinite integral

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Indefinite integrals are written in the form of `int f(x) dx = F(x) +C`

 where: `f(x)` as the integrand

           `F(x)` as the anti-derivative function 

           `C `  as the arbitrary constant known as constant of integration

For the given problem `int...

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Indefinite integrals are written in the form of `int f(x) dx = F(x) +C`

 where: `f(x)` as the integrand

           `F(x)` as the anti-derivative function 

           `C `  as the arbitrary constant known as constant of integration

For the given problem `int sec^2(x/2)tan(x/2) dx ` has a integrand in the form of a trigonometric function.

To evaluate this, we may apply u-substitution by letting `u = tan(x/2)` .

 Then, the derivative of `u` is:

`du = sec^2(x/2) *(1/2) dx`

 Rearrange this into `2 du= sec^2(x/2) dx` .

Plug-in the values on the `int sec^2(x/2)tan(x/2) dx `  , we get:

`int sec^2(x/2)tan(x/2) dx =int u *2 du`

 Apply the basic properties of integration: `int c*f(x) dx= c int f(x) dx` .

`int u *2 du =2int u du`  

Apply the Power Rule for integration:`int (x^n) dx = x^(n+1)/ (n+1) +C` .

`2int u du =2* u^(1+1)/(1+1) +C`

               `= 2 *u^2/2+C`

               `= u^2 +C`

Plug-in `u = tan(x/2)` on `u^2 +C` , we get the indefinite integral as:

`int sec^2(x/2)tan(x/2) dx =(tan(x/2))^2 +C`  or `tan^2(x/2) +C`

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