help integration (1 Viewer)

[CriminalCrab]

Please help solve. The sheet they come from didn't have answers =(

Integrate:
1. (2x+3)/sqrt(x^2+2X+3)
2. (5t^2+3)/t(t^2+1)
3. sqwiggle thing root 3 and -1 (definite integral): X^2/sqrt(4-x^2)
4. (3x^2 -8x^3-19x-14)/(x-3)(x^2+2)

thanks people

 

[Mr Slick]

I know how to do em' but cbf doing here .... lol sorry mate
btw u can check answers on wolfram

 

[Drongoski]

Q1

 

[nightweaver066]

2.

 

[Nooblet94]

<a href="http://www.codecogs.com/eqnedit.php?latex=I=\int^{\sqrt{3}}_{-1}\frac{x^2}{\sqrt{4-x^2}}\\ $Let $x=2\sin \theta\\ dx=2\cos \theta d\theta\\ x=-1\Rightarrow \theta = -\frac{\pi}{6}\\ x=3 \Rightarrow \theta = \frac{\pi}{3}\\ ~\\ I=4\int^{\frac{\pi}{3}}_{-\frac{\pi}{6}} \frac{\sin^2\theta\cos \theta d\theta}{\cos \theta}\\ =2\int^{\frac{\pi}{3}}_{-\frac{\pi}{6}} (1-\cos 2\theta)d\theta\\ =2\left [\theta-\frac{1}{2}\sin 2\theta \right ]^{\frac{\pi}{3}}_{-\frac{\pi}{6}}\\ =2\left ( \frac{\pi}{3}-\frac{\sqrt{3}}{4}@plus;\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right )\\ =\pi-\sqrt{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?I=\int^{\sqrt{3}}_{-1}\frac{x^2}{\sqrt{4-x^2}}\\ $Let $x=2\sin \theta\\ dx=2\cos \theta d\theta\\ x=-1\Rightarrow \theta = -\frac{\pi}{6}\\ x=3 \Rightarrow \theta = \frac{\pi}{3}\\ ~\\ I=4\int^{\frac{\pi}{3}}_{-\frac{\pi}{6}} \frac{\sin^2\theta\cos \theta d\theta}{\cos \theta}\\ =2\int^{\frac{\pi}{3}}_{-\frac{\pi}{6}} (1-\cos 2\theta)d\theta\\ =2\left [\theta-\frac{1}{2}\sin 2\theta \right ]^{\frac{\pi}{3}}_{-\frac{\pi}{6}}\\ =2\left ( \frac{\pi}{3}-\frac{\sqrt{3}}{4}+\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right )\\ =\pi-\sqrt{3}" title="I=\int^{\sqrt{3}}_{-1}\frac{x^2}{\sqrt{4-x^2}}\\ $Let $x=2\sin \theta\\ dx=2\cos \theta d\theta\\ x=-1\Rightarrow \theta = -\frac{\pi}{6}\\ x=3 \Rightarrow \theta = \frac{\pi}{3}\\ ~\\ I=4\int^{\frac{\pi}{3}}_{-\frac{\pi}{6}} \frac{\sin^2\theta\cos \theta d\theta}{\cos \theta}\\ =2\int^{\frac{\pi}{3}}_{-\frac{\pi}{6}} (1-\cos 2\theta)d\theta\\ =2\left [\theta-\frac{1}{2}\sin 2\theta \right ]^{\frac{\pi}{3}}_{-\frac{\pi}{6}}\\ =2\left ( \frac{\pi}{3}-\frac{\sqrt{3}}{4}+\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right )\\ =\pi-\sqrt{3}" /></a>

 

[juantheron]

(4)

 

[Carrotsticks]

I know how to do em' but cbf doing here .... lol sorry mate
btw u can check answers on wolfram

Then why bother replying?

This is what I love about BOS... somebody posting a question then some of the most talented Mathematicians on here answer 1 question each, each with their own different style.

 

[wilsondw]

2.

For the last line, I believe it should be 3ln(t)+ln(t^2+1) +c....not minus.....correct me if im wrong

 

[nightweaver066]

For the last line, I believe it should be 3ln(t)+ln(t^2+1) +c....not minus.....correct me if im wrong

Yep lol. Thanks