Harder Integration methods (1 Viewer)

[rsingh]

Oh thank you.
And while we're on the subject, does inverse trig. usually pop us in integration by substitution questions?

 

[rsingh]

Oh and I was reading one of the maths summaries, and I found this note on integration:
arcsinx +C = -arccosx + D

What does that mean?

 

[velox]

Slide covered that in one of the above posts.

 

[Slidey]

Oh and I was reading one of the maths summaries, and I found this note on integration:
arcsinx +C = -arccosx + D

What does that mean?

Hmm. Be careful. That's not really true.

Short explanation:
Derivative of inverse sin is the negative of the derivative of inverse cos.
This does NOT mean that arccosx=-arcsinx. However, it is true that if you were to integrate 1/sqrt(1-x^2), both arcsinx and -arccosx would be suitable answers. That's because it has to do with area and such, though. It's not identity. The identity is arcsinx=arccos(sqrt(1-x^2))

Long derivation:
let y=arcsinx
siny=x
dx/dy=cosy=cos(arcsinx)
So we have a triangle with a right-angle and an opposite side length of x, a hypotenuse side length of 1. Thus the adjacent side length is sqrt(1-x^2), thus arcsinx=arccos(sqrt(1-x^2))
And dx/dy=cosy=cos(arcsinx)=sqrt(1-x^2)
dy/dx=1/dx/dy, so
dy/dx=d(arcsinx)/dx=1/sqrt(1-x^2)

Now for cos:
let y=arccosx
cosy=x
dx/dy=-siny=-sin(arccosx)
Triangle: opposite side length is sqrt(1-x^2)
arccosx=arcsin(sqrt(1-x^2))
And so dx/dy=-sqrt(1-x^2)
d(arccosx)/dx=-1/sqrt(1-x^2)

Each is the negative of the other.

When did i say i'd disprove it? And i was talking about say if you had - 1/sqrt(1-x^2) and you had to integrate it and your answer was -arcsin then it would be frowned upon.

Sorry. I misunderstood you. I suppose they'd prefer you to eliminate the negative sign.

 

[velox]

Yeh, my head teacher said it pisses him off when he says that. One other crazy teacher makes you lose marks for that.