Further Pure 2
General Integration Techniques
- When an integrand involves sqrt(1-cosx) or sqrt(1+cosx), use the following identies:
cosx = 1 - 2sin²(x/2) = 2cos²(x/2) - 1
These should help you get rid of the square root.
e.g. sqrt(1-cosx) = sqrt[1-1+2sin²(x/2)] = sqrt(2) sin(x/2), which you can now integrate easily.
- For integrands of forms similar to:
1/(acosx + bsinx), 1/(a+bsinx) or 1/(a+bcosx) (This includes rsecx.)
The substitution t=tan(x/2) can make things simpler. Use the following identities to apply the substitution correctly:
sec²(x/2) = 1 + tan²(x/2) = 1 + t²
sinx = 2sin(x/2)cos(x/2) = 2tan(x/2)/sec²(x/2) = 2t/(1+t²)
cosx = cos²(x/2) - sin²(x/2) = (1 - tan²(x/2))/sec²(x/2) = (1-t²)/(1+t²)
e.g.
∫ 1/(1+2sinx) dx
= ∫ 1/[1+4t/(1+t²)] . 2/(1+t²) dt
= ∫ 2/(t²+4t+1) dt
= 2 ∫ 1/[(t+2)² - 3] dt
= 1/sqrt[3] ln|(t+2-sqrt[3])/(t+2+sqrt[3])|
Don't forget to back-substitute t=tan(x/2) in your final answer.
- Spotting "function and its derivative"-type integrands helps quite a bit.
f(x) . f'(x) dx = (1/n) [f(x)]n + C
f'(x)/f(x) dx = ln|f(x)| + C
e.g. ∫ x/(x²+1) dx = (1/2) ∫ 2x/(x²+1) dx = (1/2) ln|x²+1| + C, since x²+1 is the function and 2x is its derivative.
e.g. ∫ cos²x sinx dx = - ∫ cos²x(-sinx) dx = -(1/3)cos³x + C
- Integrands that contain fractions may require long divison. This happens when the numerator is greater than the denumerator.
e.g. x/(x-2) = 1 + 2/(x-2), which you can now integrate easily.
Reduction Formulae Techniques
- Adding zero (+1-1) might help.
e.g.
xn/(xn + 1)
= (xn + 1 - 1)/(xn + 1)
= (xn + 1)/(xn + 1) - 1/(xn + 1)
= 1 - 1/(xn + 1)
- Splitting up indices might make the integrand easier to work with.
- Sometimes you have to use integration by parts twice.
e.g. ∫ xn sinx dx
First time: u=xn, dv/dx=sinx
Second time: u=xn-1, dv/dx=cosx
General Coordinate Geometry Techniques
- To find the locus of a point as some variable (e.g. t) varies:
1. Write down the coordinates of the point, e.g. (2t, at²)
2. Express the x-coord & y-coord seperately, e.g. x=2t, y=at²
3. Make the variable the subject of one of your equations (it helps if it's the simpler one), e.g. t=x/2
4. Plug this in the other equation to get the locus of the point's path, e.g. y=at²=a(x/2)² => 4y=ax²
Sometimes you might need to use some trig identities.
e.g. Find the locus of P (3cost, 5/sint) as t varies.
x = 3cost => cost = x/3
y = 5/sint => sint = 5/y
cos²t + sin²t = 1
(x/3)² + (5/y)² = 1
And simplify if need be.
e.g. Find the locus of Q (acos2u, bsinu) as u varies.
x = acos2u => cos2u = x/a
y = bsinu => sinu = y/b
cos2u = 1 - 2sin²u
x/a = 1 - 2(y/b)²
etc.
Use your imagination and make sure you know your identities.
- You don't always have to solve a quadratic equation to use its roots.
If an equation of the form x²+ax+b=0 has roots p and q, then p+q=-a and pq=b.
e.g. Suppose you have a question that's asking you for the coordinates of the midpoint of a line joining two points on a parabola.
Let's say that the equations of the line and the parabola are y=x+c and y²=ax, respectively.
Now we want to solve these two equations simultaneously to find the coordinates of those points.
(x+c)²=ax
x²+2xc+c²=ax
x²+(2c-a)x+c²=0
Finding the roots p & q of this equation will prove to be a horrendous task. So instead, use the fact that:
p + q = -(2c-a) = a-2c
Similarly, we can find r+t, where r & t are the y-intercepts of the points.
Now all we have to do is plug this into the formula for the midpoint:
[(p+q)/2, (r+t)/2]
And we're done.
- Know what form of coordinates you should use.
For example, dealing with a certain ellipse question you might find using coordinates of the form (p, q) to suit your needs perfectly; some other times, however, you're going to have to use the parametric coordinates (acost, bsint).
The same applies to all the other conics.
Intrinsic coordinates:
1: Converting intrinsic form [s=f(w)] to cartesian form [y=g(x)].
Differentiate to give (ds/dw)=f'(w) so ds=f'(w) dw
(dy/ds)=sinw
∫1 dy = ∫sinw ds
y=∫sinw.f'(w) dw
Integrate to give a relationship between y and w. Remember not to neglect the arbitrary constant.
(dx/ds)=cosw
∫1 dx = ∫cosw ds
x=∫cosw.f'(w) dw
Integrate to give a relationship between x and w. Remember not to neglect the arbitrary constant.
Eliminate the parameter 'w' to give the cartesian relationship between y and x. Typically you'll use a trigonometric identity or make w the subject in the two equations to eliminate w.
2: Converting cartesian form [y=f(x)] to intrinsic form [s=g(w)]
y=f(x) so (dy/dx)=f'(x)=tanw
s=∫[1+(dy/dx)^2]^0.5 dx with lower limit a(where arc length is measured from) and upper limit x.
s=∫[1+(f(x))^2]^0.5 dx with lower limit a and upper limit x.
Perform the integration to find the relation s=h(x) and eliminate x using (dy/dx)=tanw, to form a relation involving only s and w.
You can convert from x=f(y) or parametric equations x=f(t), y=g(t) in a similar manner using the appropriate arc length formula.
by Atharva Inamdar
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