Question

ind_int xcos(5x) dx
u=x
du=dx
dv=cos(5x) dx
v=u-sub
{u=5x du=5
v=ind_int cos(5x)
=sin(u) du
=sin(5x)*5
(1/5)v=sin(5x)

My algebra is retricebackwards. Can someone please help me clear this up so I can better understand what I'm doing. Especially with notation and whatnot. thanks.

Answer 1

can u write this with equation forum

Answer 2

\[\int\limits_{}^{}xcos(5x)dx\]

Answer 3

My v should equal (1/5)sin(5x), but I'm unclear on the notations and it makes the algebra confusing.

Answer 4

v=\[\int\limits_{}^{}\cos(5x)\]

Answer 5

v=\[\int\limits_{}^{}\cos(5x)dx\]

Answer 6

5x=u
d(5x)/dx=du/dx
5=du/dx
du=5dx

Answer 7

\[\int xcos(5x)dx\]\[=\frac{1}{5}\int xd(sin5x)\]Integration by parts\[=\frac{1}{5}[xsin5x-\int sin5xd(x)]\]\[=\frac{1}{5}[xsin5x-(-\frac{1}{5}cos5x)]+C\]\[=\frac{1}{5}xsin5x+\frac{1}{25}cos5x+C\]Not sure if it is correct

Answer 8

therefore cos5xdx=(1/5)(5cos5x)dx=(1/5)(cos5x) 5dx=(1/5cosu)du

Answer 9

That's the right ans, but I have problems with the notation in my u-sub. There's u's, du's, v's and dx's, and when I'm solving it out algebraically the meanings of the notations get lost to me. It's hard to convey across a forum.

Answer 10

if you want i could send you some scanned texts to your email

Answer 11

Do yo understand the first step in my answer?

Answer 12

@Chumku I think you can upload the scanned text here via the ''attach file'' function.

Answer 13

ya sure

Answer 14

but i will be able to attach the file around after three hours or so