Integral using trig identities and IBP

In this video, I showed how to modify and evaluate an integral using trig identities and IBP

[jan-willemreens]

... I hope you're having a nice weekend Newton, After watching your presentation closely I've got a few comments for you: 1) At about time 3:05 ... the derivative of sin(2x) must be 2cos(2x) in your thoughts ... 2) At about time 6:11 u = x and du = 1dx = dx, dv = sin(2x)dx and v = (-1/2)cos(2x) ... 3) At about time 10:04 ... you must start by writing INT(xsin(x)cos(x))dx = (1/2)INT(xsin(2x))dx = ... etc. etc. instead of starting with (1/2)INT(xsin(x)cos(x))dx ... but you didn't make further mistakes in your following calculations luckily ... Last but not least Newton, your enthusiasm while teaching is unmatchable! Thank you again for the inspiration you spread ... Jan-W

[ferrislkpokpa]

If you were my maths teacher in college, I would have loved maths. I am now learning calculus all over again from you. Thanks do much!

[vincentmudimeli]

Mr Newton everytime I watch your video I am left my Jaws dropped you make maths funny and enjoyable.

[ikbr]

Thank you so much my understanding is so much better now! I'm smiling right now 😂

[Ni]

I hate mixing x and 2x, it's simple and half of the time (of course I mean most of the time) I screw it up. So I did this -
∫ x ½sin(2x) dx
u = 2x, ½u = x, ½du = dx
∫ ½u ½sinu ½du
⅛ ∫ u sinu du
D... I
u sinu
-1 -cosu
0 -sinu
⅛(-u cosu + sinu)
⅛(-2x cos(2x) + sin(2x)) + c

It's not a perfect world or maybe not the best way but it works for me. Loving your videos, missing the old intro but life is about change. 👍

[tajpa]

You made 2 mistakes: du=dx and it is (1/2)*integrate xsin(2x)

[ferrislkpokpa]

Your writing is so small, please make it more visible!

[trelosyiaellinika]

Try applying
Igl(uvdw) = uvw - Igl(uwdv) - Igl(vwdu). where cosxdx=dw, x=u & sinx=v, you'll get
I = xsin^2(x) - I - Igl(sin^2(x)) and eventually
I = (1/2)xsin^2(x) - x/4 + (1/8)sin(2x), which is the same, only faster.