integration problemo (1 Viewer)

[freaking_out]

use the result I{0->a}f(x)dx=I{0->a}f(a-x)dx show that:

I{0->pi/2}cos^3x/(sin^3x+cos^3x)dx=

I{0->pi/2}sin^3x/(sin^3x+cos^3x)dx

and hence evaluate the given integral.

this question is from arnold/cambridge pg. 173 q.2b

anyway, i have done the first bit, but how do i integrate that integral, let alone evaluate that integral?

 

[ND]

Hint: If I = I{0->pi/2}cos^3x/(sin^3x+cos^3x)dx and I also = I{0->pi/2}sin^3x/(sin^3x+cos^3x)dx, then 2*I = ?

but how do i integrate that integral, let alone evaluate that integral?

All it's asking you to do is find the integral, that's what evaluate means in this question.

 

[wogboy]

Yep when you have integrals of this form, just find the averages of the two expressions:

In this case find the average of:

I{0->pi/2}cos^3x/(sin^3x+cos^3x)dx

and

I{0->pi/2}sin^3x/(sin^3x+cos^3x)dx

is

1/2 * {I{0->pi/2}cos^3x/(sin^3x+cos^3x)dx + I{0->pi/2}sin^3x/(sin^3x+cos^3x)dx}
= 1/2 {I{0->pi/2}cos^3x/(sin^3x+cos^3x) + sin^3x/(sin^3x+cos^3x)dx}
= 1/2 {I{0->pi/2} (cos^3x + sin^3x)/(cos^3x + sin^3x) dx
= 1/2 {I{0->pi/2} 1 dx
= pi/4

Because you know that if a=b then the average of a and b (i.e. (a+b)/2) is equal to a and b, for any values of a and b.

 

[freaking_out]

wow, this average stuff is cool, i never thought about manipulating the integrals this way. anyway i wanted to know that if its possible to integrate the functions them selves????

 

[wogboy]

Examples like this show that it is possible to work out definite integrals (usually easily), without having to solve the indefinite integral first and then subbing & subtracting the limits. It's probably too hard to solve that integral as it is, you could try substitutions or something like that, but I wouldn't bother (I got Maple, a computer maths program, to solve that integral and I got something horribly complicated back)

 

[freaking_out]

Examples like this show that it is possible to work out definite integrals (usually easily), without having to solve the indefinite integral first and then subbing & subtracting the limits

umm.. i know, but its very hard to recognise the fact that you can use the shortcuts in a particular question.

anyway, where can i get that intergration program?

 

[ND]

In that question, you don't have to recognise it, the question tells you.

You can get maple 7 off kazaa.

edit: in a similar question, if it didn't have the cubes, you could do it using the t formulas, but this method is still much better.

 

[freaking_out]

i would really like to see the systematic way the program uses to intergrate functions, coz i always thought that integration was something that you build into yourself as instinctively, if you know what i mean...

 

[wogboy]

You can download Maple (my version is Maple 8) from Kazza, or you can use this online integrator:

http://integrals.wolfram.com/

 

[freaking_out]

yeah, the online one did not work when i plugged in the above question as an indefinite integral...anyways what answer did maple give u?

 

[wogboy]

I cos^3(x)/(sin^3(x) + cos^3(x)) dx

= (1/2)*ln[tan(x/2)^2 + 1] + (1/6)*ln[tan(x/2)^2 - 2tan(x/2) - 1] - (1/3)*ln[tan(x/2)^4 + 2tan(x/2)^3 + 2tan(x/2)^2 - 2tan(x/2) + 1] + x/2 + C

{Note that tan(x/2)^n = tan((x/2)^n), not (tan(x/2))^n }

That's the answer from Maple. The online integrator I showed you above actually does work for that question it's just that your input isn't correct (in fact it gives a better & more simplified answer than Maple). Click the "How to enter your input" button for instructions. Now that you have the answer for:

I cos^3(x)/(sin^3(x) + cos^3(x))

As an exercise, find:

I sin^3(x)/(sin^3(x) + cos^3(x))

 

[freaking_out]

As an exercise, find:I sin^3(x)/(sin^3(x) + cos^3(x))

:mad1:

 

[wogboy]

aww come on it's easy!! :gridnod:

Hint: Use a similar technique as solving the definite integral, with the averages/adding stuff.