Question

Integral

Answer 1

Integrate:
\[\int\limits \sqrt{ax^2+bx+c}.dx\]

Answer 2

I am thinking to complete the square, then use a trig sub

Answer 3

1. complete the sqaure

Answer 4

square

Answer 5

This is my approach:
Pull out the a:
\[\sqrt{a}\int\limits \sqrt{x^2+\frac{bx}{a}+\frac{c}{a}}dx\]
Complete the square:
\[\sqrt{a} \int\limits \sqrt{ \left( x + \frac{b}{2a} \right)^2 + \frac{c}{a}-\frac{b^2}{4a^2}}dx\]
Call
\[\beta=\frac{c}{a}-\frac{b^2}{4a^2}; \xi=x+\frac{b}{2a} \implies d \xi =dx\]
\[\sqrt{a}\int\limits \sqrt{\xi^2+\beta}d \xi\]
Now make the substitution:
\[\tan(\eta)=\frac{\xi}{\beta} \implies \sec^2(\eta) d \eta = \frac{d \xi}{\beta} \rightarrow \beta \sec^2(\eta) d \eta=d \xi; \]
\[\sec(\eta)=\frac{\sqrt{\xi^2+\beta^2}}{\beta} \rightarrow \beta \sec(\eta)=\sqrt{\xi^2+\beta^2}\]
\[\sqrt{a}\beta^2 \int\limits \sec^3(\eta)d \eta=\sqrt{a}\beta^2 \int\limits \left[ \tan^2(\eta)\sec(\eta)+\sec^2(\eta)\right] d \eta\]

Answer 6

yes, that's exactly what I did (without all the greek letters lol)

Answer 7

From here you can of course do integration by parts:
\[\phi=\tan(\eta) \rightarrow d \phi = \sec^2(\eta)d \eta; d \rho =\sec(\eta)\tan(\eta) d \eta \rightarrow \rho=\sec(\eta)\]
\[\sqrt{a}\beta^2 \left[ \tan(\eta)\sec(\eta)-\int\limits \sec(\eta)\sec^2(\eta)\right]\]
So:
\[\int\limits \sec^3(\eta)=\tan(\eta)\sec(\eta)-\int\limits \sec^3(\eta)d \eta +\int\limits \sec^2(\eta) deta \implies \int\limits \sec^3(\eta)=\]
\[\frac{1}{2}\tan(\eta)\sec(\eta)+\frac{1}{2}\int\limits \sec^2(\eta)d \eta\]
Therefore we arrive at:
\[\frac{\sqrt{a}\beta^2}{2} \tan(\eta) \left[ \sec(\eta)+1\right] \rightarrow \frac{\sqrt{a} \xi}{2}\sqrt{\xi^2+\beta^2}=\frac{\sqrt{a} (x+\frac{b}{2a})}{2}\sqrt{(x+\frac{b}{2a})^2+(\frac{c}{a}-\frac{b^2}{4a^2})^2}+C\]

Answer 8

The end of that is just whatever beta is equal to up top^^ I hope you guys like my quick disposition, sorry it took so long I was calculating and typing this on the fly :D

Answer 9

good show :D

Answer 10

Integrals <3 hahahahaha

Answer 11

No one is posting good questions so I spiced it up!

Answer 12

@malevolence19 since that is the case here are a couple to rack your head over ;)
Hard:
1)\[\large \int_{1}^{\infty} e^{-x^2}(\frac1x-2x\ln x)dx\]2)\[\large \int_{-\frac\pi2}^{\frac\pi2}\sinh x\cos^2x-\frac{\cos x}{\cos x\sin x}\]Harder:
3)prove that\[\large\int_{0}^{\pi}\frac{x\sin x}{1+\cos^2x}dx=\frac{\pi^2}{4}\]4)\[\large\int_{0}^{1}\frac{\ln(x+1)}xdx\]good luck, happy integrating

Answer 13

I know that 3) is a u-sub on cos(x) and then it turns into a geometric series under assumption that cos(x)<1

Answer 14

perhaps, I couldn't solve that one
Zarkon did, but he didn't do it that way
The answer to that can be found under my profile or in meta-math

Answer 15

Can I post us a question to pick your brain?

Answer 16

Unfortunately I have to go eat, but next time for sure!

Answer 17

Message me when you are back then!

Answer 18

will do :D