Starting with this definite integral:
$$\huge \int\limits_{0}^{1} \frac{r^3}{\sqrt{25+r^2}} dr$$

After having trouble trying to find an appropriate u and dv to use with integration by parts, I noticed the denominator sure looked a lot like the Pythagorean Theorem when you're solving for C. So also tried
$$C=\sqrt{25+r^2}$$
$$r=\pm\sqrt{C^2-25}$$
$$r^3=(C^2-25)^{\frac{3}{2}}$$
Still no luck... I get:
$$=\frac{r^4\sqrt{25+r^2}}{4}-\int\limits_{0}^{1}-\frac{r^3}{4}(25+r^2)^{-3/2} dr$$
Which is ugh... worse.

(as always will give a medal to the most helpful)

Question

Starting with this definite integral:
$$\huge \int\limits_{0}^{1} \frac{r^3}{\sqrt{25+r^2}} dr$$

After having trouble trying to find an appropriate u and dv to use with integration by parts, I noticed the denominator sure looked a lot like the Pythagorean Theorem when you're solving for C.  So also tried
$$C=\sqrt{25+r^2}$$
$$r=\pm\sqrt{C^2-25}$$
$$r^3=(C^2-25)^{\frac{3}{2}}$$
Still no luck... I get:
$$=\frac{r^4\sqrt{25+r^2}}{4}-\int\limits_{0}^{1}-\frac{r^3}{4}(25+r^2)^{-3/2} dr$$
Which is ugh... worse.

(as always will give a medal to the most helpful)

agent47 · Answer

Use the trig substitution

agent47 · Answer

No i think it's Sin5t or something

anonymous · Answer

*confused*

agent47 · Answer

Put u=25tanx

anonymous · Answer

u=5 tanx

agent47 · Answer

Yes lol finally we got it right, mukushla is correct

agent47 · Answer

http://www.sosmath.com/calculus/integration/trigsub/trigsub.html

anonymous · Answer

i mean r=5*tanx

anonymous · Answer

using the process that  tan x --> sec^2 x when you take the derivative, but 5tanx?  Isn't that just making things worse?

anonymous · Answer

|dw:1340377293898:dw| 
it will be easier now..