Integration by Parts: Troubleshooting the Formula for x cos 5x dx

lemurs
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kay I am having difficulties with this..
Knowing the gerneral formuala of

|uv'=uv- |vu'

i using a nonehomework question i was trying to make sure i had it down pat was having problems..

| x cos 5x dx

but for some reason i don't get the right answer when it done...
If I have u=x, du=1
and
v'=cos 5x dx

v= 1/5 sin5x? or did i crew up some where I been having trouble here ..

so i can do the substion and all but this stuff is screwing with help please.
 
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\int x\cos 5x = \frac{x}{5}\sin 5x - \frac{1}{5}\int \sin 5x dx.

\int x\cos 5x = \frac{x}{5}\sin 5x +\frac{1}{25}\cos 5x

So \int udv = uv-\int vdu.

\int \sin 5x = -\frac{\cos 5x}{5}.
 
Last edited:
kay my major problem is that 1/5 where does it come from.

how does cos 5x dx = 1/5 sin 5x..
 
when you take the derivative of 1/5 sin 5x, you get cos 5x by doing the chain rule...you have to take the derivative of the argument because it is more complex than just an x
 
t6hanks Maps Think i undersand it now... hopefully the homework will be easier now
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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