`int sec^3 (pix) dx` Find the indefinite integral

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marizi eNotes educator| Certified Educator

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  integral problem `int sec^3(pix) dx` , we may evaluate this using u-substitution.

Let: `u = pix` then `du = pi dx` or  `(du)/pi =dx` .

The integral becomes:

`int sec^3(pix) dx =int sec^3(u) * (du)/pi`

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

`int sec^3(u) * (du)/pi =1/piint sec^3(u) du`

Apply integration formula for secant function:

`int sec^n(x) dx = (sec^(n-1)(x)sin(x))/(n-1) + (n-2)/(n-1) int sec^(n-2)(x) dx`

We get:

`1/piint sec^3(u) du =1/pi [(sec^(3-1)(u)sin(u))/(3-1) + (3-2)/(3-1) int sec^(3-2)(u) du]`

        `=1/pi [(sec^2(u)sin(u))/(2) + (1)/(2) int sec^(1)(u) du]`

For the integral of  `int sec^(1)(u) du` or  `int sec^(u) du` , we may apply `int sec(theta) d theta = ln(sec(theta)+tan(theta))+C` .

Then,`int sec^(u) du =ln(sec(u)+tan(u))+C`

The complete indefinite integral will be:

`1/piint sec^3(u) du =1/pi [(sec^2(u)sin(u))/(3-1) + (1)/(2) int sec^(1)(u) du]`

           `=1/pi [(sec^2(u)sin(u))/(2) + (1)/(2)[ln(sec(u)+tan(u))]]+C`

           `=1/pi [(sec^2(u)sin(u))/2+ln(sec(u)+tan(u))/2]+C`

           `=(sec^2(u)sin(u))/(2pi)+ln(sec(u)+tan(u))/(2pi)+C`

          or  `tan(u)/(2picos(u)) + 1/(2pi)ln((1+sin(u))/cos(u))+C`

Plug-in `u = pix` , we get the final indefinite integral as:

`int sec^3(pix) dx=(sec^2(pix)sin(pix))/(2pi)+ln(sec(pix)+tan(pix))/(2pi)+C`

                        or `tan(pix)/(2picos(pix)) + 1/(2pi)ln((1+sin(pix))/cos(pix))+C`

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