Question

How do you find the indefinite integral of `sqrt(25 + x^2)`?

Answer 1

`1/2(xsqrt(25+x^2)+25lnabs(x+sqrt(25+x^2)))+C`

Explanation:

Use trigonometric substitution.

Draw a triangle with an angle `theta`.

Label the opposite side as `x` and adjacent side as `5`.

Now examine the triangle and notice that

`tantheta=x/5`

`rArrx=5tantheta`

`rArrdx=5sec^2theta` `d` `theta`

By the Pythagorean Theorem, the hypotenuse of the triangle is:

`sqrt(25+x^2)`

So we can write

`sectheta=sqrt(25+x^2)/5`

`rArrsqrt(25+x^2)=5sectheta`

Let's now rewrite the integral in terms of `theta`

`intsqrt(25+x^2)dxrArrint(5sectheta)(5sec^2theta` `d` `theta)`

`rArr25int` `sec^3theta` `d` `theta`

Integrating `sec^3theta` can be tricky. There is a shortcut formula, but I'll do it the long way for teaching purposes.

Separate `(sec^3theta` `d` `theta)` into `(sectheta*sec^2theta` `d` `theta)` and use integration by parts.

Let:

`u=sectheta`
`du=secthetatantheta` `d` `theta`

`dv=sec^2theta` `d` `theta`
`v=tantheta`

Then:

`rArrint` `sec^3theta` `d` `theta=secthetatantheta-int` `tan^2thetasectheta` `d` `theta`

`rArrint` `sec^3theta` `d` `theta=secthetatantheta-int` `(sec^2theta-1)sectheta` `d` `theta`

`rArrint` `sec^3theta` `d` `theta=secthetatantheta-int` `(sec^3theta-sectheta)` `d` `theta`

`rArr2int` `sec^3theta` `d` `theta=secthetatantheta+int` `sectheta` `d` `theta`

`rArr2int` `sec^3theta` `d` `theta=secthetatantheta+lnabs(sectheta+tantheta)`

`rArrint` `sec^3theta` `d` `theta=1/2(secthetatantheta+lnabs(sectheta+tantheta))+C`

And let's not forget that our integral was multiplied by 25!

`rArr25int` `sec^3theta` `d` `theta=25/2(secthetatantheta+lnabs(sectheta+tantheta))+C`

Now put everything back in terms of `x`

`rArr25/2(sqrt(25+x^2)/5x/5+lnabs(sqrt(25+x^2)/5+x/5))+C`

`rArr1/2(xsqrt(25+x^2)+25lnabs((sqrt(25+x^2)+x)/5))+C`

`rArr1/2(xsqrt(25+x^2)+25lnabs(x+sqrt(25+x^2))-25ln5)+C`

Then we can absorb the constant `-(25ln5)/2` into `C` to get our final answer:

`rArr1/2(xsqrt(25+x^2)+25lnabs(x+sqrt(25+x^2)))+C`