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integrals
- Thread starter jeca86
- Start date
It is not necessary to reference a table for this problem.
Let \(\displaystyle \L u=x^2+1,\) so that \(\displaystyle \L du=2x\,dx.\)
Then:
\(\displaystyle \L \qquad\begin{eqnarray}
\int x^5\sqrt{x^2+1}\,dx
&=&\frac12\int(x^2+1-1)^2\sqrt{x^2+1}(2x\,dx)\\
&=&\frac12\int(u-1)^2\sqrt{u}\,du\\
&=&\frac12\int(u^{5/2}-2u^{3/2}+u^{1/2})\,du
\end{eqnarray}\)
Let \(\displaystyle \L u=x^2+1,\) so that \(\displaystyle \L du=2x\,dx.\)
Then:
\(\displaystyle \L \qquad\begin{eqnarray}
\int x^5\sqrt{x^2+1}\,dx
&=&\frac12\int(x^2+1-1)^2\sqrt{x^2+1}(2x\,dx)\\
&=&\frac12\int(u-1)^2\sqrt{u}\,du\\
&=&\frac12\int(u^{5/2}-2u^{3/2}+u^{1/2})\,du
\end{eqnarray}\)
The 2 was insterted inside the integral. In order to keep the integral equivilant, you need to put a 1/2 outside.
For example, if you wanted to integrate INT[cos(3x)dx], you are missing a 3 inside, so you put the 3 in, and put a 1/3 outside:
1/3 INT[cos(3x)*3dx] = 1/3 INT[cos(u)*du] = 1/3*sin(3x)
For your second question, the x^5 was changed into:
(x^2)^2 * x
The last x was needed for the du on the end, and the inside x^2 was rewritten as x^2+1-1. Notice that the "+1-1" =0. This was done, it seems, to make substitution easier.
So: x^5 = (x^2+1-1)^2*x. And since u = x^2 + 1, it changed it to: (u-1)^2*x
The last x was moved to the end of the integral and attatched to the 2 that was put in and the dx to make du. See: (2xdx) in their example.
It wasn't a dumb question, by the way. No offense to the poster, but there was a huge assumption on their part to think you'd just 'get it'. More steps should have been shown in my opinion. Maybe its that I am a student and get how fusturating it can be to read a texbook and run into a few nicely skipped steps. I dunno...
For example, if you wanted to integrate INT[cos(3x)dx], you are missing a 3 inside, so you put the 3 in, and put a 1/3 outside:
1/3 INT[cos(3x)*3dx] = 1/3 INT[cos(u)*du] = 1/3*sin(3x)
For your second question, the x^5 was changed into:
(x^2)^2 * x
The last x was needed for the du on the end, and the inside x^2 was rewritten as x^2+1-1. Notice that the "+1-1" =0. This was done, it seems, to make substitution easier.
So: x^5 = (x^2+1-1)^2*x. And since u = x^2 + 1, it changed it to: (u-1)^2*x
The last x was moved to the end of the integral and attatched to the 2 that was put in and the dx to make du. See: (2xdx) in their example.
It wasn't a dumb question, by the way. No offense to the poster, but there was a huge assumption on their part to think you'd just 'get it'. More steps should have been shown in my opinion. Maybe its that I am a student and get how fusturating it can be to read a texbook and run into a few nicely skipped steps. I dunno...