... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

But this is wrapped inside the legs-uncrossed version of the product rule. (See previous.)

Re: Calculus Trig Integration/substitution, by parts, improper integrals?

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 =/

Mar 23rd 2013, 07:57 AM

tom@ballooncalculus

Re: Calculus Trig Integration/substitution, by parts, improper integrals?

Ok I do usually show less in the first picture, which makes the direction of travel a bit clearer. Which is, roughly...

... 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).

Use integration by parts with $\displaystyle u = \ln{(x)} \implies du = \frac{1}{x}\,dx$ and $\displaystyle \displaystyle dv = \frac{1}{\sqrt{x}}\,dx \implies v = 2\sqrt{x} $ and the integral becomes

Re: Calculus Trig Integration/substitution, by parts, improper integrals?

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

Mar 23rd 2013, 12:37 PM

goku900

Re: Calculus Trig Integration/substitution, by parts, improper integrals?

Why did you change the boundaries to 0 to 1 on this problem

lower bound 0 upper bound pi/4 ∫ sec^4xtan^4x dx

Mar 23rd 2013, 04:43 PM

Prove It

Re: Calculus Trig Integration/substitution, by parts, improper integrals?

Quote:

Originally Posted by goku900

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.

Quote:

Originally Posted by goku900

Why did you change the boundaries to 0 to 1 on this problem

lower bound 0 upper bound pi/4 ∫ sec^4xtan^4x dx

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.

Mar 23rd 2013, 05:05 PM

hollywood

Re: Calculus Trig Integration/substitution, by parts, improper integrals?

The integral of $\displaystyle \csc{x}$ has a "trick" shortcut method:

and since $\displaystyle \frac{d}{dx}\csc{x}=-\cot{x}\csc{x}$ and $\displaystyle \frac{d}{dx}\cot{x}=-\csc^2{x}$, the numerator is the derivative of the denominator, so