another quick inverse trig question (1 Viewer)

[kooltrainer]

can you integrate sin inverse x?
how?

 

[powerdrive]

do u do 4 unit? use integration by parts

 

[ssglain]

Since integration by parts has been mentioned, I will briefly explain it and show you how to integrate sin<SUP>-1</SUP>x using IBP.

Basically, IBP works by the reverse of the product rule used in differentiation, so the concept should be fairly easy to understand even if you don't do MX2.

Now, the product rule says d/dx (uv) = u'v + uv'.
Then integrating both sides of the equation gives:
∫[d/dx (uv)] = ∫u'v + ∫uv'
uv = ∫u'v + ∫uv'
.: ∫uv' = uv - ∫u'v
(I omitted the "dx" at the end of each integral for the sake of clarity.)

So, if you encounter something that cannot be integrated using other standard procedures, such as ∫sin<SUP>-1</SUP>x dx, ∫xlnx dx, ∫xe<SUP>x</SUP> dx, etc., break up the integrand into two parts u and v'. Take care to select u as the part that is easily differentiable and v' as the part that can be easily integrated to produce a relatively simple object.

In our case, the integrand can only be broken up into sin<SUP>-1</SUP>x and 1.
Let u = sin<SUP>-1</SUP>x and v' = 1 (note that it would be a futile attempt to do this the other way around)
Then u' = 1/√(1 - x²) and v = x

Then using our neat little formula:
∫sin<SUP>-1</SUP>x dx
= xsin<SUP>-1</SUP>x - ∫x/√(1 - x²) dx
= xsin<SUP>-1</SUP>x + (1/2)*∫-2x(1 - x²)<SUP>-1/2 </SUP>dx
= xsin<SUP>-1</SUP>x + (1/2)*[2(1 - x²)<SUP>1/2</SUP>] + C
= xsin<SUP>-1</SUP>x + √(1 - x²) + C

Alternatively, if this was a definite integral that you need to evaluate you can always sketch y = sin<SUP>-1</SUP>x and flip your graph and deal with rectangle - ∫sin y dy using limits taken from the y-axis.

 

[powerdrive]

wow how long did it take you to write that post, with all the integration symbols. i did pretty much the same thing, except i hand wrote it, was about to scan and post it up, wouldve been much easier.

if you are doing definite integration, integrating inverse sin x means finding the area under this curve. the area under inverse sin x = area of rectangle - area under sin x. (look at 4U 2005 HSC q 5.c)

 

[ssglain]

wow how long did it take you to write that post, with all the integration symbols. i did pretty much the same thing, except i hand wrote it, was about to scan and post it up, wouldve been much easier.

if you are doing definite integration, integrating inverse sin x means finding the area under this curve. the area under inverse sin x = area of rectangle - area under sin x. (look at 4U 2005 HSC q 5.c)

Yeah, I also thought about writing it by hand and scanning but this didn't take too long. Although there are a few typos, now that I've taken the time to read it over.

 

[kooltrainer]

Since integration by parts has been mentioned, I will briefly explain it and show you how to integrate sin<SUP>-1</SUP>x using IBP.

Basically, IBP works by the reverse of the product rule used in differentiation, so the concept should be fairly easy to understand even if you don't do MX2.

Now, the product rule says d/dx (uv) = u'v + uv'.
Then integrating both sides of the equation gives:
∫[d/dx (uv)] = ∫u'v + ∫uv'
uv = ∫u'v + ∫uv'
.: ∫uv' = uv - ∫u'v
(I omitted the "dx" at the end of each integral for the sake of clarity.)

So, if you encounter something that cannot be integrated using other standard procedures, such as ∫sin<SUP>-1</SUP>x dx, ∫xlnx dx, ∫xe<SUP>x</SUP> dx, etc., break up the integrand into two parts u and v'. Take care to select u as the part that is easily differentiable and v' as the part that can be easily integrated to produce a relatively simple object.

In our case, the integrand can only be broken up into sin<SUP>-1</SUP>x and 1.
Let u = sin<SUP>-1</SUP>x and v' = 1 (note that it would be a futile attempt to do this the other way around)
Then u' = 1/√(1 - x²) and v = x

Then using our neat little formula:
∫sin<SUP>-1</SUP>x dx
= xsin<SUP>-1</SUP>x - ∫x/√(1 - x²) dx
= xsin<SUP>-1</SUP>x + (1/2)*∫-2x(1 - x²)<SUP>-1/2 </SUP>dx
= xsin<SUP>-1</SUP>x + (1/2)*[2(1 - x²)<SUP>1/2</SUP>] + C
= xsin<SUP>-1</SUP>x + √(1 - x²) + C

Alternatively, if this was a definite integral that you need to evaluate you can always sketch y = sin<SUP>-1</SUP>x and flip your graph and deal with rectangle - ∫sin y dy using limits taken from the y-axis.

is that part of 4unit maths? or is it extension1?

 

[SoulSearcher]

is that part of 4unit maths? or is it extension1?

Extension 2 Maths.

 

[ssglain]

Well, see the only MX2 part was knowing ∫uv' = uv - ∫u'v. Once you hit ∫sin<SUP>-1</SUP>x dx = xsin<SUP>-1</SUP>x - ∫x/√(1 - x²) dx, everything else that follows relies on no more than MX1 integration.

 

[kooltrainer]

hi ssglain, do u no james wang.. he said he knows u .. aha

 

[ssglain]

hi ssglain, do u no james wang.. he said he knows u .. aha

Yes, lol.