Inverse trig integration (1 Viewer)

eyeseeyou · Aug 7, 2016

Help

∫▒〖dt/√(3-12t-18t^2 ) 〗

trecex1 · Aug 7, 2016

factorise 18 outside root , rearrange the rest to complete the square and you get a standard asin integral.

eyeseeyou · Aug 7, 2016

∫▒〖(1+x)dx/(1+x^2 ) 〗

I don't know how to approach this question

eyeseeyou · Aug 7, 2016

trecex1 said:
factorise 18 outside root , rearrange the rest to complete the square and you get a standard asin integral.

I found that confusing

InteGrand · Aug 7, 2016

eyeseeyou said:
Help

∫▒〖dt/√(3-12t-18t^2 ) 〗

$\noindent Call the integral $I$. Complete the square in the denominator. We have$

$$\begin{align*}3-12t -18t^2 &= -18\left(t^2 + \frac{2}{3}t -\frac{1}{6}\right) \\ &= -18\left[ \left(t+\frac{1}{3}\right)^2 -\frac{1}{9}-\frac{1}{6}\right] \\ &= -18\left(t+\frac{1}{3}\right)^2 + 5.\end{align*}$$

$\noindent Hence using the fact that $\sqrt{18} = 3\sqrt{2}$, a combination of inspection and reverse chain rule gives$

$\begin{align*}I &= \frac{1}{3\sqrt{2}} \sin ^{-1} \left(\frac{3\sqrt{2}\left(t+\frac{1}{3}\right)}{ \sqrt{5}}\right) + c.\end{align*}$

You can easily simplify the argument of the inverse sine if you want.

trecex1 · Aug 7, 2016

eyeseeyou said:
I found that confusing

eyeseeyou said:
∫▒〖(1+x)dx/(1+x^2 ) 〗

I don't know how to approach this question

you can just split that up into 2 integrals and you'll get atanx + 1/2 ln(1+x^2) + c

eyeseeyou · Aug 7, 2016

trecex1 said:
you can just split that up into 2 integrals and you'll get atanx + 1/2 ln(1+x^2) + c

Thanks

eyeseeyou · Aug 7, 2016

How do I do this:∫_0^(-1)▒〖dt/√(1+6t-3t^2 ) 〗

My working:

∫_0^(-1)▒〖dt/√(-3t^2+6t+1) 〗

∫_0^1▒〖dt/√(-3(t^2-2t-1/3) ) 〗

∫_0^1▒〖dt/√(-3(t-1)^2-1-1/3) 〗

integral95 · Aug 7, 2016

I can't comprehend the limit from your typing, but the integral goes like this.

$\int \frac{dt}{\sqrt{1+6t-3t^2}} = \\ \\ \int \frac{1}{\sqrt{3}} \frac{dt}{\sqrt{\frac{4}{3} - (t-1)^2}} = \frac{1}{\sqrt{3}} sin^{-1}(\frac{\sqrt{3}}{2}(t-1))$

eyeseeyou · Aug 7, 2016

integral95 said:
I can't comprehend the limit from your typing, but the integral goes like this.

$\int \frac{dt}{\sqrt{1+6t-3t^2}} = \\ \\ \int \frac{1}{\sqrt{3}} \frac{dt}{\sqrt{\frac{4}{3} - (t-1)^2}} = \frac{1}{\sqrt{3}} \frac{\sqrt{3}}{2} sin^{-1}(\frac{\sqrt{3}}{2}(t-1))$

borders are -1 and 0

leehuan · Aug 7, 2016

eyeseeyou said:
borders are -1 and 0

All you need is there you can sub in the boundaries yourself.

eyeseeyou · Aug 27, 2016

I am kinda confused on the working to this question

1)Find the area btwn the curve y=sin^-1 x, x=0, x=1 and the x-axis

Could someone please show me the working out with explanations. I do not understand the theory behind it

Thanks

eyeseeyou · Aug 27, 2016

BTW I read the textbook and am hardly understanding what it's talking about

InteGrand · Aug 27, 2016

eyeseeyou said:
I am kinda confused on the working to this question

1)Find the area btwn the curve y=sin^-1 x, x=0, x=1 and the x-axis

Could someone please show me the working out with explanations. I do not understand the theory behind it

Thanks

$\noindent It's just asking for the area under the graph of $y=\sin ^{-1} x$ between $x=0$ and $1$. So the answer is $A = \int _{0} ^{1} \sin ^{-1}x \, \mathrm{d}x$. One way to compute this integral is via \textsl{integration by parts}. This is beyond HSC 3U though. The 3U way to do this question is to note that the area desired is $A_2 -A_1$, where $A_2$ is the area of the rectangle bounded by $x=0$, $x=1$, $y=0$ and $y= \frac{\pi}{2}$, and $A_1$ is the area between the curve $y=\sin^{-1}x $, $y=0$ and $y=\frac{\pi}{2}$ (draw diagram to see it).$

$\noindent Thus $A = A_2 -A_1 = \frac{\pi}{2} - \int _{0} ^{\frac{\pi}{2}} \sin y \, \mathrm{d}y = \frac{\pi}{2} -1$.$

eyeseeyou · Aug 27, 2016

BTW integrand how do I do integration borders on latex?

eyeseeyou · Aug 27, 2016

Integrand this is what I don't understand
It says you must use the rectangle method since we can't integrate straight away

$\int _{0} ^{\frac{\pi}{2}} x \, dy$
$\int _{0} ^{\frac{\pi}{2}} siny \, dy$
-cosy (borders from pi/2 and 0)
-[0-1]
1

InteGrand · Aug 27, 2016

eyeseeyou said:
Integrand this is what I don't understand
It says you must use the rectangle method since we can't integrate straight away

$\int _{0} ^{\frac{\pi}{2}} x \, dy$
$\int _{0} ^{\frac{\pi}{2}} siny \, dy$
-cosy (borders from pi/2 and 0)
-[0-1]
1

Yeah I did that method if you see my working. In your diagram, the red area needs to be subtracted from the overall rectangle's area in order to give us the black area (which is what we're after). The rectangle has area pi/2 and the red area is 1, so the answer is pi/2 – 1.

eyeseeyou · Aug 27, 2016

InteGrand said:
Yeah I did that method if you see my working. In your diagram, the red area needs to be subtracted from the overall rectangle's area in order to give us the black area (which is what we're after). The rectangle has area pi/2 and the red area is 1, so the answer is pi/2 – 1.

But why is it "the area of red is the integral of pi/2 to 0 x dy

Were we integrating wrt x axis or y axis?

InteGrand · Aug 27, 2016

eyeseeyou said:
But why is it "the area of red is the integral of pi/2 to 0 x dy

Were we integrating wrt x axis or y axis?

$\noindent For the red area, it's like a normal integral except the curve we're finding the area `under' is of the form $x = f(y)$ (so the roles of $x$ and $y$ are reversed compared to when you find the area under a typical curve, like bounded by the $x$-axis). (Here, $f(y)=x = \sin y$, since $y = \sin ^{-1} x$ on the curve.) Since the area in question is bounded by the $y$-axis, we integrate this function of $y$, $f(y)$, between the bounds for $y$ $\Big{(}$which are $y=0$ to $y =\frac{\pi}{2}$$\Big{)}$. This is like how it this were an area bounded under a function $h(x)$'s graph and the $x$-axis, we'd integrate the function $h(x)$ between the $x$ limits. So all that's happened is we've swapped the usual roles of $x$ and $y$.$

eyeseeyou · Aug 27, 2016

InteGrand said:
$\noindent For the red area, it's like a normal integral except the curve we're finding the area `under' is of the form $x = f(y)$ (so the roles of $x$ and $y$ are reversed compared to when you find the area under a typical curve, like bounded by the $x$-axis). (Here, $f(y)=x = \sin y$, since $y = \sin ^{-1} y$ on the curve.) Since the area in question is bounded by the $y$-axis, we integrate this function of $y$, $f(y)$, between the bounds for $y$ $\Big{(}$which are $y=0$ to $y =\frac{\pi}{2}$$\Big{)}$. This is like how it this were an area bounded under a function $h(x)$'s graph and the $x$-axis, we'd integrate the function $h(x)$ between the $x$ limits. So all that's happened is we've swapped the usual roles of $x$ and $y$.$

Would the same principle apply for another question like "find the area btwn y=sin^-1 x/2, x=1 and x=2 and the x-axis?

Inverse trig integration (1 Viewer)

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