Inverse trig integration (1 Viewer)

trecex1

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

eyeseeyou

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∫▒〖(1+x)dx/(1+x^2 ) 〗

I don't know how to approach this question :(
 

eyeseeyou

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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

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


 
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eyeseeyou

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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

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

InteGrand

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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


 
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eyeseeyou

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



-cosy (borders from pi/2 and 0)
-[0-1]
1
 

InteGrand

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



-cosy (borders from pi/2 and 0)
-[0-1]
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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

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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?
 

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