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C4 Integration by substitution problem

Here's the question:

Use the substitution u = tanx to find the value of:

∫(1 + tan²x) / (1 - tan²x) dx between the limits π/6 and 0

I do this and my new integral is:

∫1 / (1 - u²) du between the limits of 1/√3 and 0

The question says that you should give your answer in logarithmic form but I just don't know how to solve this new integral since i can't introduce new variables on the top to make it the bottom differentiated!

Any help is much appreciated.
Thanks in advance,
RoosterDude

Reply 1

Speleo

u = tanx
du = sec^2 dx
but 1 + tan^2x = sec^2 x
so du = 1 + u^2 dx

∫ (1 + tan^2x/1 - tan^2x) dx [pi/6 - 0]

∫ (1/1-u^2) du [1/√3 - 0]

Partial fractions: 1/1-u^2 -> 0.5/1-u + 0.5/1+u

(1/2) ∫ 1/1-u + 1/1+u du [1/√3 - 0]

= 1/2 [-ln(1-u) + ln(1+u)] [1/√3 - 0]
= 1/2 [-ln(1-1/√3) + ln(1+1/√3)] - [0 + 0]

= 1/2 ln [(1+1/√3)/(1-1/√3)]

= ln[(1+1/√3)/(1-1/√3)^1/2]

Could probably be simplified.

Reply 2

bob_54321

it's a standard integral, look in a formula booklet.

∫1 / (1 - u²) = ( ln ((1+u)/(1-u))) / 2

(it says give it in log form because you can also give it in terms of inverse tanh.)

Reply 3

bob_54321

or you can do it that way but if it is in an exam you can just quote the result from the formulae booklet (or your memory!).

Reply 4

Speleo

I hadn't even heard of that, what are the other standard integrals?

All I know is the trigonometric functions etc. and ∫ f'(x)/f(x) dx = ln[f(x)].

Reply 5

bob_54321

well, in my formulae booklet it gives integrals for 1/(a^2 + x^2), 1/(a^2 - x^2), 1/(x^2 - a^2) and then the same with the denominator square rooted. they are mainly just the trig and hyperbolic ones but there are a couple of other forms of some of them that involve logs. you don't need to know any, you can work them all out reasonably easily, just saves a bit of time in exams and the like if you know them (or look them up).

Reply 6

Speleo

Oh, ok, thanks :smile:

Reply 7

RoosterDude

That's great! Thanks a lot. I can't believe I didn't see that!

So here's one more thing i don't understand:

It says: The diagram shows the circle x² + y² = a² and the line x = a/2. Find the area of the shaded region. (Diagram attatched below)

Now if you make y the subject you get y = √(a² - x²) and so I figured that the area of the shaded region would be the integral:

∫√(a² - x²) dx between a and a/2

However in the answers it says that it is two times the above integral. Why is this?

Thanks again.

Circle Problem.bmp223.8KB

Reply 8

Speleo

You are getting problems with half the shape being under the x-axis.

Reply 9

RoosterDude

Speleo

You are getting problems with half the shape being under the x-axis.

I understand what you're saying but if this is the case, wouldn't the integral come out to be zero since there are equal areas above and below the axis as the circle has centre, the origin?

Reply 10

Speleo

I understand what you're saying but if this is the case, wouldn't the integral come out to be zero since there are equal areas above and below the axis as the circle has centre, the origin?

It might well come out to zero, have you checked?

Reply 11

yusufu

RoosterDude

I understand what you're saying but if this is the case, wouldn't the integral come out to be zero since there are equal areas above and below the axis as the circle has centre, the origin?

The function y = √(a² - x²) only describes the curve for y >= 0. The part below the x axis comes from the negative root, i.e. y = -√(a² - x²).

Reply 12

evariste_3086

RoosterDude

I understand what you're saying but if this is the case, wouldn't the integral come out to be zero since there are equal areas above and below the axis as the circle has centre, the origin?

for it to be a function you only consider top half of the curve.
Integration will give the area of top half.
By symmetry the bottom bit is the same as the top half.
total area is 2 lots of area worked out by the integration.

Reply 13

Speleo

Ah of course, there are two different square roots.
Thanks YYYY :smile:

Reply 14

RoosterDude

evariste

for it to be a function you only consider top half of the curve.
Integration will give the area of top half.

Is this only true since it is the root function then? Because the integral of x^3 between 3 and -3 would give 0 yes?

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