Solving Tricky Integration Problems: Tips and Examples for Success

[Haths]

I have three questions, I don't need a full working through, but I'd prefer some hints or a simmilar example for where I am going wrong/need help. The answers arn't important, but the method of working them out is.

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Homework Statement

A line AB is given;

Integ{ x dy - y dx }

Where the line is the peramitised set of equations: x=t² and y=t+1 between 0<t<1

2. My attempt at a solution

I've assumed that this means that I am intergrating;

Integ{ t² [d/dy] - t+1 [d/dx] } dt

Which is;

Integ{ -t²+2t } dt ==> 1/3 t³ + t² |¹₀

Therefore: 1.333...

I can't remember if this is what you do or not and considering I was going to ask some other questions I thought I'd ask this one too.

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Homework Statement

Knowing the double angle formula for sine(a+b). Intergrate;

Integ{ sin(2x)cos(3x) } dx

The Attempt at a Solution

My first thought was simply rearrage the double angle formula for the integrals;

Integ{ sin(5x) } dx - Integ{ cos(2x)sin(3x) } dx

But of course that gets me nowhere. Then I thought, what if I convert sin(2x) to 2sin(x)cos(x) and make my integral;

2*Integ{ sin(x)cos(x)cos(3x) } dx

But again that doesn't appear to help as I can't use a change of varible on the cos(3x) term, and if I want to expand that using the same double angle rules as before you get a nasty;

Integ{ 2sin(x)cos⁴(x) - 2sin³(x)cos²(x) - 4sin^3(x)cos²(x) }

Which I suppose is doable, but there should be a trick for this problem, and that's what I'm missing

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Homework Statement

Knowing d/du tan(u) = 1 + tan²(u)

Intergrate;

Integ{ 1 / (1 + x²) } dx

THEN find;

Integ^Infinity₀{ 1 / (3 + 2x²) }

The Attempt at a Solution

For that last part I haven't a clue...

The first part however I assumed that I could let x = tan(u) and so change my integral to;

Integ{ 1 / tan(u) } du

Which using the product rule I believe intergrates to cos(u) Ln| sin(u) | + 1 as 1/tan = cos/sin

But I don't know is that's the right method as ignoring the 'hint' it gets nasty quickly :(.

Cheers and thanks in advance,
Haths

 

[Thaakisfox]

In the second one, the trick is to use the angle addition formula. I will right them down under each other and leave you to figure out why this is good ;) :

\sin{(a+b)}=\sin a \cdot \cos b + \sin b \cdot \cos a
\sin{(a-b)}=\sin a \cdot \cos b - \sin b \cdot \cos a

Now add the to equations together ;))
This same "trick" is used in deriving the Fourier coefficients... :D

In the third one, you did the right substitution, just work it out:

x=\tan u \Longrightarrow \frac{dx}{du}=\frac{d}{du}\tan u \Longrightarrow dx= (1+\tan^2u)du

So the integral:

\int\frac{dx}{1+x^2}=\int\frac{(1+\tan^2u)du}{1+\tan^2u}=\int du = u+C=\text{arctan}(x)+C

 

[Haths]

While I admire your help in not giving me the answers 'off the bat' so that I can work at this futher, I'm still not that much closer to understanding how to do the 2nd one:

The explanation was akin to my first thoughts, but looking back the only thing that struck me was such a nasty trick I throwed my pencil at the wall...

sin(2x)cos(3x) = 1/2 sin(5x)

+++
1/2 sin(2x+3x) = 1/2 (2sin(2x)cos(3x))
+++

Therefore;

-1/10 cos(5x) is the integral

I wasn't impressed as the double angle formula doesn't even need mentioning, it was a red herring in some aspects...

Anyhow thanks for the help, but some questions when answered breed more questions;

THEN find;

IntegInfinity0{ 1 / (3 + 2x2) }

Using the same trick I get down to;

Integ{ 1 + tan²(u) / 1 + 2(1+tan²(u) } du

From;

Integ{ 1 + tan²(u) / 3 + 2tan²(u) } du

Which then stumps me in rearrangment so that I can eliminate the 1+tan²(u) term.

Say it was 2 + ... on the denominator, that would be simple, you could factorise except for that the fact it will be complex... and unless my basic algerbra is failing me, I'm not sure where I'm going wrong :(.

Haths

 

[Thaakisfox]

Yeah, that happens to me sometimes also, throwing away my pen... :D

But the problem is, that your expression is incorrect :(

using the angle addition formula:

\sin(5x)=\sin(2x+3x)=\sin(2x)\cos(3x)+\sin(3x)\cos(2x) \neq 2\sin(2x)\cos(3x)

But we have to use, the trick I mentioned:

\sin{(2x+3x)}=\sin (2x) \cdot \cos (3x) + \sin (3x) \cdot \cos (2x)
\sin{(2x-3x)}=\sin (2x) \cdot \cos (3x) - \sin (3x) \cdot \cos (2x)

Adding these two equations together:

\sin{(2x+3x)}+\sin{(2x-3x)}= 2\sin (2x) \cdot \cos (3x)

Hence we obtain:
\sin (2x) \cdot \cos (3x)=\frac{\sin{(2x+3x)}+\sin{(2x-3x)}}{2}=\frac{\sin(5x)+\sin(-x)}{2}=\frac{\sin(5x)-\sin(x)}{2}

Now this can be integrated easily..


For the second one, consider an integral of the form:

\int\frac{1}{a+bx^2}\;dx

Where a,b are some positive real numbers. (In your case a=3, b=2).

Now let's do the following, manipulations:

\int\frac{1}{a+bx^2}\;dx = \int\frac{1}{a\left(1+\frac{b}{a}x^2\right)}\;dx= \int\frac{1}{a\left(1+\left(\frac{\sqrt{b}}{\sqrt{a}}x\right)^2\right)}\;dx=\frac{1}{a}\int\frac{1}{1+\left(\frac{\sqrt{b}}{\sqrt{a}}x\right)^2}\;dx

Here put: \tan(u)=\frac{\sqrt{b}}{\sqrt{a}}x
Then we have:
(1+\tan^2(u))du=\frac{\sqrt{b}}{\sqrt{a}}dx \Longrightarrow dx=\frac{\sqrt{a}}{\sqrt{b}}(1+\tan^2(u))du

So our integral:

\frac{1}{a}\int\frac{1}{1+\left(\frac{\sqrt{b}}{\sqrt{a}}x\right)^2}\;dx = \frac{1}{a}\int\frac{\frac{\sqrt{a}}{\sqrt{b}}(1+\tan^2(u))}{1+\tan^2(u)};du=\frac{\sqrt{a}}{a\sqrt{b}}\int du = \frac{1}{\sqrt{ab}}u+C = \frac{1}{\sqrt{ab}} \text{arctan}\left(x\sqrt{\frac{b}{a}}\right)+C

So in your case the integral:

\int_0^{\infty}\frac{1}{3+2x^2};dx = \left[\frac{1}{\sqrt{6}}\text{arctan}\left(x\sqrt{\frac{2}{3}}\right)\right]_0^{\infty}=\frac{\pi}{2\sqrt{6}}

since we know that: \lim_{m\to \infty}\text{arctan}(m) = \frac{\pi}{2} and \text{arctan}(0)=0

Always when you have integral of this type, try to manipulate so that you have 1+\dots in the denominator :D