How do you integrate #int x/sqrt(x^2+1)# by trigonometric substitution?

Question

Evelyn Agard · Accepted Answer

#sqrt(x^2 + 1) + C#

Let #x= tantheta#. Then #dx = sec^2thetad theta#

#=> int tan theta/sqrt((tan^2theta + 1)) sec^2theta d theta#
#=> int tantheta/sqrt(sec^2theta) sec^2theta d theta#
#=> int tantheta/sectheta sec^2theta d theta#
#=> int tan theta sec theta d theta#

This is a common integral--#int(tanxsecx)dx = secx + C#.

#=> sec theta + C#

We now draw an imaginary triangle.

The definition of #sectheta# is #"hypotenuse"/("side adjacent" theta)# because #secx = 1/cosx#. In this image, #sectheta = sqrt(x^2 + 1)#.

Therefore, the integral can be simplified to #sqrt(x^2 + 1) + C#.

Hopefully this helps!

Emilia Aguilar · Answer

By inspection rather than trig substitution.

Notice that the #x# in the top is proportional to the derivative of the function under the square root. So just write down #(x^2+1)^(-1/2+1)# as a trial (rather like integrating any power of #x#) and differentiate it. By the chain rule, the function under the square root will provide the #2x# and the #2#'s will cancel, so the trial was an immediate success. Alternatively substitute #u=x^2+1#, #dx=(du)/(2sqrt(u-1))# and the integral becomes #(1/2)int u^(-1/2)du# = #u^(1/2)+C#=#sqrt(x^2+1)+C#

Christopher Agresta · Answer

undefined To integrate $ \int \frac{x}{\sqrt{x^2 + 1}} $ using trigonometric substitution, we can substitute $ x = \tan(\theta) $, which implies $ dx = \sec^2(\theta) \, d\theta $.

After substitution:

1. $ x = \tan(\theta) $
2. $ dx = \sec^2(\theta) \, d\theta $
3. $ \sqrt{x^2 + 1} = \sqrt{\tan^2(\theta) + 1} = \sqrt{\sec^2(\theta)} = \sec(\theta) $

Now, the integral becomes:

$ \int \frac{\tan(\theta)}{\sec(\theta)} \cdot \sec^2(\theta) \, d\theta $

$ = \int \sin(\theta) \cos(\theta) \, d\theta $

Using the identity $ \sin(2\theta) = 2\sin(\theta)\cos(\theta) $:

$ = \frac{1}{2} \int \sin(2\theta) \, d\theta $

Now, integrate $ \sin(2\theta) $ with respect to $ \theta $:

$ = -\frac{1}{2} \cos(2\theta) + C $

Substituting back $ \theta = \arctan(x) $, we get:

$ = -\frac{1}{2} \cos\left(2\arctan(x)\right) + C $

This is the indefinite integral of $ \frac{x}{\sqrt{x^2 + 1}} $ using trigonometric substitution.