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Hiiii i was solving these problems and dont think theyre right at all://

 

1. ∫1/6 0 (1/(sqrt(1-9x^2))dx **I got 6.49 as my answer....I don't think that's right

 

2. ∫5 3 (1/(x^2-6x+13))dx ** I don't know how to do this one...please help!

 Mar 26, 2019
 #1
avatar+110153 
+3

∫1/6 0 (1/(sqrt(1-9x^2))dx

 

\(\displaystyle\int_0^{\frac{1}{6}}\frac{1}{\sqrt{1-9x^2}}dx\)

 

Here is a site that should help

 

https://www.integral-calculator.com/

 

That site got     pi/18

 Mar 26, 2019
 #2
avatar+110153 
+3

 

I probably do these a little differently from most people.

I cannot remember the integrals so I so them more from scratch.

 

\(\displaystyle\int_0^{\frac{1}{6}}\frac{1}{\sqrt{1-9x^2}}dx\)

 

I am going to use this triangle.

 

\(cos\theta=\sqrt{1-9x^2}\\ sin\theta=3x\\ x=\frac{sin\theta}{3}\\ \frac{dx}{d\theta}=\frac{cos\theta}{3}\\ dx=\frac{cos\theta}{3}d\theta\)

 

also

\(when \;x=0  \\ sin\theta=0\\ \theta=0\\~\\ when\;x=1/6\\ sin\theta=3*(1/6)=1/2\\ \theta = \frac{\pi}{6}\)

 

 

\(\displaystyle\int_0^{\frac{1}{6}}\frac{1}{\sqrt{1-9x^2}}dx\\ =\displaystyle\int_0^{\frac{1}{6}}\frac{1}{cos\theta}dx\\ =\displaystyle\int_0^{\frac{\pi}{6}}\frac{1}{cos\theta}\frac{cos\theta}{3}d\theta\\ =\displaystyle\int_0^{\frac{\pi}{6}}\frac{1}{3}d\theta\\ =\left[\frac{\theta}{3}\right]_0^{\frac{\pi}{6}}\\ =\frac{\pi}{18} \)

 Mar 26, 2019
 #3
avatar+110153 
+1

deleted

 Mar 26, 2019
edited by Melody  Mar 26, 2019
 #4
avatar+110153 
+3

 ∫5 3 (1/(x^2-6x+13))dx

 

\(\displaystyle\int_3^{5}\frac{1}{\sqrt{x^2-6x+13}}dx\\ =\displaystyle\int_3^{5}\frac{1}{\sqrt{x^2-6x+9+4}}dx\\ =\displaystyle\int_3^{5}\frac{1}{\sqrt{2^2+(x-3)^2}}dx\\\)

 

 

I did this one in a very similar fashion.

I used this triangle

 

 

See if you can take it from there   laugh

 Mar 26, 2019
 #5
avatar+110153 
+1

If anyone wants to show guest how to do it in a more traditional way.

That is fine :)

 Mar 26, 2019
 #6
avatar+25481 
+2

1. 

∫1/6 0 (1/(sqrt(1-9x^2))dx

\(\large{\int \limits_{0}^{\frac{1}{6}} \dfrac{1}{ \sqrt{1-9x^2} }dx}\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{ \large{\int \limits_{0}^{\frac{1}{6}} \dfrac{1}{ \sqrt{1-9x^2} }\ dx} } \\\\ &=& \displaystyle \int \limits_{0}^{\dfrac{1}{6}} \dfrac{1}{ \sqrt{1-(3x)^2} }\ dx \\\\ && \quad \boxed{ \text{substitute: } \\ u = 3x,\ du=3\ dx,\\ x=0 \Rightarrow u=3\cdot 0=0,\\ x=\dfrac{1}{6}\Rightarrow u=3\cdot \dfrac{1}{6}=\dfrac{1}{2} } \\\\ &=& \displaystyle \int \limits_{0}^{\dfrac{1}{2}} \dfrac{1}{ \sqrt{1-u^2} } \dfrac{du}{3} \\\\ &=& \dfrac{1}{3}\displaystyle \int \limits_{0}^{\dfrac{1}{2}} \dfrac{1}{ \sqrt{1-u^2} }\ du \\\\ && \quad \boxed{ \text{substitute: } \\ u = \sin(\theta),\ du=\cos(\theta)d\theta,\\ u=0 \Rightarrow \theta=\arcsin(0)=0,\\ u=\dfrac{1}{2}\Rightarrow \theta=\arcsin\left(\dfrac{1}{2}\right)=\dfrac{\pi}{6}} \\\\ &=& \dfrac{1}{3}\displaystyle \int \limits_{0}^{\dfrac{\pi}{6}} \dfrac{\cos(\theta)}{ \sqrt{1-\Big(\sin(\theta)\Big)^2} }\ d\theta \\\\ &=& \dfrac{1}{3}\displaystyle \int \limits_{0}^{\dfrac{\pi}{6}} \dfrac{\cos(\theta)}{ \cos(\theta) }\ d\theta \\\\ &=& \dfrac{1}{3}\displaystyle \int \limits_{0}^{\dfrac{\pi}{6}} d\theta \\\\ &=& \dfrac{1}{3} \Big[ \theta \Big]_{0}^{\frac{\pi}{6}} \\\\ &=& \dfrac{1}{3}\cdot \dfrac{\pi}{6} \\\\ &\mathbf{=}& \mathbf{\dfrac{\pi}{18}} \\ \hline \end{array}\)

 

laugh

 Mar 26, 2019
edited by heureka  Mar 26, 2019
edited by heureka  Mar 26, 2019
edited by heureka  Mar 26, 2019
 #7
avatar+25481 
+1

2. 

∫5 3 (1/(x^2-6x+13))dx

\(\large{\int \limits_{3}^{5} \dfrac{1}{ \sqrt{x^2-6x+13} }\ dx}\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{ \displaystyle \int \limits_{3}^{5} \dfrac{1}{ \sqrt{x^2-6x+13} }\ dx} \\\\ &=& \displaystyle \int \limits_{3}^{5} \dfrac{1}{ \sqrt{(x-3)^2-9+13} }\ dx \\\\ &=& \displaystyle \int \limits_{3}^{5} \dfrac{1}{ \sqrt{(x-3)^2+4} }\ dx \\\\ &=& \displaystyle \int \limits_{3}^{5} \dfrac{1} { \sqrt{ 4 \left( 1+ \dfrac{ (x-3)^2 } {4} \right) } }\ dx \\\\ &=& \displaystyle \int \limits_{3}^{5} \dfrac{1} { 2 \sqrt{ 1+ \left(\dfrac{x-3}{2}\right)^2 } }\ dx \\\\ &=& \dfrac{1}{2} \displaystyle \int \limits_{3}^{5} \dfrac{1} {\sqrt{ 1+ \left(\dfrac{x-3}{2}\right)^2 } }\ dx \\\\ && \quad \boxed{ \text{substitute: } \\ \dfrac{x-3}{2} = \sinh(\theta),\ \dfrac{1}{2}dx=\cosh(\theta)d\theta,\ dx=2\cosh(\theta)d\theta \\ x=3 \Rightarrow \theta=\sinh^{-1}\left(\dfrac{3-3}{2}\right)=0,\\ x=5\Rightarrow \theta=\sinh^{-1}\left(\dfrac{5-3}{2}\right)\\ =\sinh^{-1}(1)=\ln(1+\sqrt{1^2+1})=0.88137358702} \\\\ &=& \dfrac{1}{2} \displaystyle \int \limits_{0}^{0.88137358702} \dfrac{1} {\sqrt{ 1+ \Big(\sinh(\theta)\Big)^2 } }2\cosh(\theta)\ d\theta \\\\ &=& \displaystyle \int \limits_{0}^{0.88137358702} \dfrac{\cosh(\theta)} { \cosh(\theta) }\ d\theta \\\\ &=& \displaystyle \int \limits_{0}^{0.88137358702} d\theta \\\\ &=& \Big[ \theta \Big]_{0}^{0.88137358702} \\\\ &=& 0.88137358702 \\\\ && \mathbf{ \displaystyle \int \limits_{3}^{5} \dfrac{1}{ \sqrt{x^2-6x+13} }\ dx = \sinh^{-1}(1) = 0.88137358702} \\ \hline \end{array}\)

 

 

laugh

 Mar 26, 2019
edited by heureka  Mar 26, 2019

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