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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Yoda on May 12, 2014, 04:53:57 pm

Title: Show that question for integration?
Post by: Yoda on May 12, 2014, 04:53:57 pm: Making use of the substitution x=sin(Ɵ)

Show that ∫1/sqrt(9-9x^2)dx = ∫ 1/3dƟ
Title: Re: Show that question for integration?
Post by: b^3 on May 12, 2014, 05:08:18 pm: $\begin{alignedat}{1}\int\frac{1}{\sqrt{9-9x^{2}}}dx \\ \text{Let }x=\sin\left(\theta\right) \\ \frac{dx}{d\theta} & =\cos\left(\theta\right) \\ \int\frac{1}{\sqrt{9-9x^{2}}}dx & =\int\frac{1}{\sqrt{9-9\sin^{2}\left(x\right)}}dx \\ & =\int\frac{1}{3}\frac{1}{\sqrt{1-\sin^{2}\left(x\right)}}dx \\ \text{Replace }dx\text{ with }\cos\left(x\right)d\theta \\ \text{As }\sin^{2}\left(x\right)+\cos^{2}\left(x\right) & =1 \\ \int\frac{1}{\sqrt{9-9x^{2}}}dx & =\int\frac{1}{3}\frac{1}{\sqrt{\cos^{2}\left(x\right)}}\cos\left(x\right)d\theta \\ & =\int\frac{1}{3}\frac{1}{\cos\left(x\right)}\cos\left(x\right)d\theta \\ & =\int\frac{1}{3}d\theta \end{alignedat}$

Technically shouldn't show the $x$ variable in there when we have $d\theta$ , but since the $\cos(\theta)$ cancels out it doesn't really matter.
Title: Re: Show that question for integration?
Post by: Zealous on May 12, 2014, 05:11:58 pm: Quote from: Yoda on May 12, 2014, 04:53:57 pm
Making use of the substitution x=sin(Ɵ)

Show that ∫1/sqrt(9-9x^2)dx = ∫ 1/3dƟ
[edit] Beaten by Mr Cubed, although we approached it in slightly different ways. :D

$\begin{eqnarray*}\\\\let\ x & = & sin(\theta)\\\theta & = & sin^{-1}(x)\\\frac{d\theta}{dx} & = & \frac{1}{\sqrt{1-x^{2}}}\\\\\int\frac{1}{\sqrt{9-9x^{2}}}dx & = & \int\frac{1}{\sqrt{9(1-x^{2})}}dx=\int\frac{1}{3}(\frac{1}{\sqrt{1-x^{2}}})dx\\sub\ in\ \frac{d\theta}{dx} & = & \frac{1}{\sqrt{1-x^{2}}}\\\int\frac{1}{\sqrt{9-9x^{2}}}dx & = & \int\frac{1}{3}(\frac{d\theta}{dx})dx=\int\frac{1}{3}d\theta\\\end{eqnarray*}\\$
Title: Re: Show that question for integration?
Post by: M_BONG on May 12, 2014, 07:23:46 pm: Quote from: b^3 on May 12, 2014, 05:08:18 pm
$\begin{alignedat}{1}\int\frac{1}{\sqrt{9-9x^{2}}}dx \\ \text{Let }x=\sin\left(\theta\right) \\ \frac{dx}{d\theta} & =\cos\left(\theta\right) \\ \int\frac{1}{\sqrt{9-9x^{2}}}dx & =\int\frac{1}{\sqrt{9-9\sin^{2}\left(x\right)}}dx \\ & =\int\frac{1}{3}\frac{1}{\sqrt{1-\sin^{2}\left(x\right)}}dx \\ \text{Replace }dx\text{ with }\cos\left(x\right)d\theta \\ \text{As }\sin^{2}\left(x\right)+\cos^{2}\left(x\right) & =1 \\ \int\frac{1}{\sqrt{9-9x^{2}}}dx & =\int\frac{1}{3}\frac{1}{\sqrt{\cos^{2}\left(x\right)}}\cos\left(x\right)d\theta \\ & =\int\frac{1}{3}\frac{1}{\cos\left(x\right)}\cos\left(x\right)d\theta \\ & =\int\frac{1}{3}d\theta \end{alignedat}$

Technically shouldn't show the $x$ variable in there when we have $d\theta$ , but since the $\cos(\theta)$ cancels out it doesn't really matter.

Just a random question, would you be able to take out the 1/3 outside of the integral (if the final answer didn't require it within the intergral?)
Title: Re: Show that question for integration?
Post by: Zealous on May 12, 2014, 07:28:27 pm: Quote from: Zezima. on May 12, 2014, 07:23:46 pm
Just a random question, would you be able to take out the 1/3 outside of the integral (if the final answer didn't require it within the intergral?)
Definitely, just remember that if you take out 1/3 you'll just have the integral of 1 and also note that we're integrating with respect to theta.