How does one use trigonometric substitution to integrate (x^3)/sqrt(x^2+25) dx?

#intx^3/sqrt(x^2+25) dx#
So I made #x=5tan(theta#) and
#dx=5sec^2(theta)#
I then plugged in for x and dx, I won't bore you with the in between, basically I made it to #5^3(u^3/3-u)+C# after taking the integral, (#u=sec(theta)# and #du=sec(theta)tan(theta)#) so I start working my way back to the x variables and I get #5^3(1/3 (sqrt(x^2+25)/5)^3-sqrt(x^2+25)/5)+C# and from here I am fairly stumped on how to bring it all around.. I have already worked out what the triangle would look like (x as opposite #theta#, 5 as adjacent to #theta# and #sqrt(x^2+25)# for the hypotenuse. Did I mess up a step? If not, where do I go from here?

1 Answer
Jan 30, 2018

# int \ (x^3)/sqrt(x^2+25) \ dx = 1/3sqrt(x^2+25)(x^2-50)+ C #

Explanation:

We seek:

# I = int \ (x^3)/sqrt(x^2+25) \ dx #

Rather than using a trigonometric substitution, a more appropriate substitution would be to let:

# u=x^2+25 iff x^2=u-25 \ \ #, and # (du)/dx =2x #

Which if we substitute into the integral, we get:

# I = 1/2 \ int \ (x^2)/sqrt(x^2+25) \ (2x) \ dx #
# \ \ = 1/2 \ int \ (u-25)/sqrt(u) \ du #
# \ \ = 1/2 \ int \ sqrt(u)-25/sqrt(u) \ du #

Which is now trivial to integrade,. and doing so we get:

# I = 1/2 \ {u^(3/2)/(3/2) -25u^(1/2)/(1/2) }+ C #
# \ \ = 1/2 \ {2/3usqrt(u) - 50sqrt(u)}+ C #
# \ \ = 1/3usqrt(u) - 25sqrt(u)+ C #
# \ \ = 1/3sqrt(u)(u - 75)+ C #

And restoration of the earlier substitution yields:

# I = 1/3sqrt(x^2+25)(x^2+25 - 75)+ C #
# \ \ = 1/3sqrt(x^2+25)(x^2-50)+ C #