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)`...
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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