Just in case a picture helps with 5, for a start... integral of cosec
) lower bound 2 upper bound 4 ∫ sqrt(9x^2-25)/ x^3 dx
2) lower bound 0 upper bound pi/4 ∫ sec^4xtan^4x dx
3) 0 to pi ∫ e^(cosx)sin(2x) dx
4) 4 to 9∫ ln(x)/sqrt(x) dx
5) Derive the basic Integral
∫csc(x)dx=ln[cot(x)-csc(x)]+C
Help on any of these would be appreciated.
Just in case a picture helps with 5, for a start... integral of cosec
That particular pic has a step-by-step - click the step-forward button.
Here's some 'lazy integration by parts'...
... where (key in spoiler) ...
Spoiler:
Here's some more...
... where (key in spoiler) ...
Spoiler:
Full size
In all cases, start from bottom left, work anti-clockwise (kinda).
Enjoy!!
PS, no.2 blows up half way along from zero to pi/4. Was it perhaps meant to be sec^4(x) tan^4(x) ?
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I'm so sorry but that way of showing integration by parts confuses me so much, without certain intermediate steps and using u and v I'm pretty lost. Take the first example you did:
So we work in a counter clockwise direction starting at the bottom left. OK. so the problem splits to lnx+1/sqrt(x) then from there a bunch of hidden intermediate steps happen and it's kinda difficult to follow how this was done out when I wasn't taught this way =/
Ok I do usually show less in the first picture, which makes the direction of travel a bit clearer. Which is, roughly...
Integration by parts (however you do it) is about working backwards through the product rule.
So you want to see the integrand as the result of a product rule differentiation.
But the result of a product rule differentiation is one product plus another. The bottom row here...
You have just the one product...
Nonetheless, you can fill out the inevitable consequences of putting your product in that place...
(only inevitable once you have chosen between legs crossed and uncrossed, i.e. which part to integrate immediately and which not)
... and then cancel those consequences out, like this, so that the bottom row is again equal to what you wanted to integrate...
And then you can complete the top row so that it differentiates to give the bottom...
... which is equal to your original product (the integrand).
That's exactly what the u and v business is doing, too, of course. It's a handy formula if you know how to use it, but if you can't square it with the above you probably haven't grasped how it works, and will get hung up on the notation (the u and v).
The csc(x) integration is tricky, thanks a bunch tho. I have questions
In the third line you make the sin in the numerator negative and also pull a negative out of the integral. Why do you need to do this?
Also how would anyone know to break up the 1/(1-u)(1+u) into 1/2 in the numerator...?!?
Yeah that negative sign in the beggining is really confusing to me, after you get the first solution with the +C, explain how that breaks up into that 1-u/1+u) term
I made it negative because when you make the substitution u = cos(x), you need to note that its derivative is du/dx = -sin(x), or du = -sin(x) dx. In order to turn it into the du integral, we need to replace -sin(x) dx with du.
The answer to your other question is to use Partial Fractions.
Remember that u = tan(x). I changed the bounds because when x = 0, u = tan(0) = 0 and when x = pi/4, u = tan(pi/4) = 1. By doing this, we avoid having to convert back to a function of x after integration, we can substitute the new u bounds straight away.