1. ## 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. $\displaystyle \int e^-^1 \cos (3t) dt$
2. $\displaystyle \int \sin^3x \sqrt {cos (x)} dx$ ----- Nevermind, I got this one
3. $\displaystyle \int^{4}_{1} \sqrt {t} \ln(t) dt$
4. $\displaystyle \int^{\pi/4}_{0} \sin (3x) \cos (x) dx$

2. Originally Posted by nirva
3. $\displaystyle \int^{4}_{1} \sqrt {t} \ln(t) dt$
Using integration by parts by allowing,
$\displaystyle u=\ln t$
$\displaystyle v'=\sqrt{t}$
Thus,
$\displaystyle u'=1/t=t^{-1}$
$\displaystyle v=2/3t^{3/2}$
Thus, we have,
$\displaystyle \frac{2}{3} \ln t \cdot t^{3/2}-\frac{2}{3}\int t^{3/2}t^{-1}dt$
Thus, the integrand is,
$\displaystyle \int t^{1/2}dt=\frac{2}{3}t^{3/2}$
Thus,
$\displaystyle \left \frac{2}{3} \ln t \cdot t^{3/2}-\frac{4}{9}t^{3/2} \right|^{4}_1 \approx 4.28$

3. Originally Posted by nirva
4. $\displaystyle \int^{\pi/4}_{0} \sin (3x) \cos (x) dx$
Use integration by parts with,
$\displaystyle u=\sin 3x$
$\displaystyle v'=\cos x$
Thus,
$\displaystyle u'=3\cos 3x$
$\displaystyle v=\sin x$
The integrand becomes,
$\displaystyle \sin 3x\sin x-3\int \cos 3x\sin xdx$
Use integration by parts again with,
$\displaystyle u=\cos 3x$
$\displaystyle v'=\sin x$
Thus,
$\displaystyle u'=-3\sin 3x$
$\displaystyle v=-\cos x$
Thus, the integrand becomes,
$\displaystyle \sin 3x\sin x-3 \left( -\cos 3x \cos x-3\int \sin 3x\cos x dx \right)$
Thus,
$\displaystyle \int \sin (3x) \cos (x) dx$=$\displaystyle \sin 3x\sin x+3\cos 3x\cos x+9\int \sin 3x \cos xdx$
Thus, (solve for this integral as if a variable)
$\displaystyle -\frac{1}{8}\left( \sin 3x\sin x+3\cos 3x\cos x \right)$
Since you know the anti-derivative you can find the definite integral.

4. 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. $\displaystyle \int e^-^1 \cos (3t) dt$
I presume you mean,
$\displaystyle \int e^{-t} \cos 3t dt$
Use integration by parts with,
$\displaystyle u'=e^{-t}$
$\displaystyle v=\cos 3t$
Thus,
$\displaystyle u=-e^{-t}$
$\displaystyle v'=-3\sin 3t$
By parts,
$\displaystyle -e^{-t}\cos 3t-3\int e^{-t}\sin 3t dt$
Use integration by parts again,
$\displaystyle u'=e^{-t}$
$\displaystyle v=\sin 3t$
Thus,
$\displaystyle u=-e^{-t}$
$\displaystyle v'=3\cos 3t$
By parts,
$\displaystyle -e^{-t}\cos 3t-3\left(-e^{-t}\sin 3t+3\int e^{-t}\cos 3t dt \right)$
Thus,
$\displaystyle \int e^{-t} \cos 3t dt$=$\displaystyle -e^{-t}\cos 3t+3e^{-t}\sin 3t-9\int e^{-t}\cos 3tdt$
Solve for integrand as if to say an equation,
$\displaystyle \int e^{-t} \cos 3t dt=\frac{1}{10}e^{-t}(3\sin 3t-\cos t)+C$

5. Originally Posted by ThePerfectHacker
I presume you mean,
$\displaystyle \int e^{-t} \cos 3t dt$
Use integration by parts with,
$\displaystyle u'=e^{-t}$
$\displaystyle v=\cos 3t$
Thus,
$\displaystyle u=-e^{-t}$
$\displaystyle v'=-3\sin 3t$
Yea it is $\displaystyle \int e^{-t} \cos 3t dt$ The t was too small, sorry
Is the lable for the integration by parts wrong? Like the u being the u'

6. Originally Posted by nirva
Yea it is $\displaystyle \int e^{-t} \cos 3t dt$ 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,
$\displaystyle e^{-t}$
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.