# Some Antideriv-integral problems

• May 20th 2006, 03:38 PM
nirva
Some Antideriv-integral problems
Find the antiderivatives - I tired working on these but I can't seem to get the correct answer for it
1. $
\int e^-^1 \cos (3t) dt
$

2. $
\int \sin^3x \sqrt {cos (x)} dx
$
----- Nevermind, I got this one
3. $
\int^{4}_{1} \sqrt {t} \ln(t) dt
$

4. $
\int^{\pi/4}_{0} \sin (3x) \cos (x) dx
$
• May 20th 2006, 06:16 PM
ThePerfectHacker
Quote:

Originally Posted by nirva
3. $
\int^{4}_{1} \sqrt {t} \ln(t) dt
$

Using integration by parts by allowing,
$u=\ln t$
$v'=\sqrt{t}$
Thus,
$u'=1/t=t^{-1}$
$v=2/3t^{3/2}$
Thus, we have,
$\frac{2}{3} \ln t \cdot t^{3/2}-\frac{2}{3}\int t^{3/2}t^{-1}dt$
Thus, the integrand is,
$\int t^{1/2}dt=\frac{2}{3}t^{3/2}$
Thus,
$\left \frac{2}{3} \ln t \cdot t^{3/2}-\frac{4}{9}t^{3/2} \right|^{4}_1 \approx 4.28$
• May 20th 2006, 06:27 PM
ThePerfectHacker
Quote:

Originally Posted by nirva
4. $
\int^{\pi/4}_{0} \sin (3x) \cos (x) dx
$

Use integration by parts with,
$u=\sin 3x$
$v'=\cos x$
Thus,
$u'=3\cos 3x$
$v=\sin x$
The integrand becomes,
$\sin 3x\sin x-3\int \cos 3x\sin xdx$
Use integration by parts again with,
$u=\cos 3x$
$v'=\sin x$
Thus,
$u'=-3\sin 3x$
$v=-\cos x$
Thus, the integrand becomes,
$\sin 3x\sin x-3 \left( -\cos 3x \cos x-3\int \sin 3x\cos x dx \right)$
Thus,
$\int \sin (3x) \cos (x) dx$= $\sin 3x\sin x+3\cos 3x\cos x+9\int \sin 3x \cos xdx$
Thus, (solve for this integral as if a variable)
$-\frac{1}{8}\left( \sin 3x\sin x+3\cos 3x\cos x \right)$
Since you know the anti-derivative you can find the definite integral.
• May 20th 2006, 07:03 PM
ThePerfectHacker
Quote:

Originally Posted by nirva
Find the antiderivatives - I tired working on these but I can't seem to get the correct answer for it
1. $
\int e^-^1 \cos (3t) dt
$

I presume you mean,
$\int e^{-t} \cos 3t dt$
Use integration by parts with,
$u'=e^{-t}$
$v=\cos 3t$
Thus,
$u=-e^{-t}$
$v'=-3\sin 3t$
By parts,
$-e^{-t}\cos 3t-3\int e^{-t}\sin 3t dt$
Use integration by parts again,
$u'=e^{-t}$
$v=\sin 3t$
Thus,
$u=-e^{-t}$
$v'=3\cos 3t$
By parts,
$-e^{-t}\cos 3t-3\left(-e^{-t}\sin 3t+3\int e^{-t}\cos 3t dt \right)$
Thus,
$\int e^{-t} \cos 3t dt$= $-e^{-t}\cos 3t+3e^{-t}\sin 3t-9\int e^{-t}\cos 3tdt$
Solve for integrand as if to say an equation,
$\int e^{-t} \cos 3t dt=\frac{1}{10}e^{-t}(3\sin 3t-\cos t)+C$
• May 20th 2006, 08:33 PM
nirva
Quote:

Originally Posted by ThePerfectHacker
I presume you mean,
$\int e^{-t} \cos 3t dt$
Use integration by parts with,
$u'=e^{-t}$
$v=\cos 3t$
Thus,
$u=-e^{-t}$
$v'=-3\sin 3t$

Yea it is $\int e^{-t} \cos 3t dt$ The t was too small, sorry
Is the lable for the integration by parts wrong? Like the u being the u'
• May 21st 2006, 09:04 AM
ThePerfectHacker
Quote:

Originally Posted by nirva
Yea it is $\int e^{-t} \cos 3t dt$ The t was too small, sorry
Is the lable for the integration by parts wrong? Like the u being the u'

Let me give you a tip for LaTeX,
If you want to write,
$e^{-t}$
I saw you did it as,
Code:

e^-^t
The more advaned way is to write,
Code:

e^{-t}
Note the { } play a role of parantheses in LaTeX.