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Can someone help me with these two problems. Thanks a lot.
1. integral of cos^3 thelta sin thelta d thelta, u= cos thelta
this is what I did so far: du/dx= -sin thelta du= -sin thelta dx => -du=sin thelta d thelta ...I dont even know if this is correct. I need to sub. u into the function but dont know how.
2. integral sec^2(1/x)/x^2 dx , u = 1/x
I got -ln x du = dx .....I dont know what to do at all
$\displaystyle \int \cos^3{t} \cdot \sin{t} \, dt$
$\displaystyle u = \cos{t}$
$\displaystyle du = -\sin{t} \, dt$
$\displaystyle -\int \cos^3{t} \cdot (-\sin{t}) \, dt$
sub ...
$\displaystyle -\int u^3 \, du$
do the integration.
$\displaystyle \int \sec^2\left(\frac{1}{x}\right) \cdot \frac{1}{x^2} \, dx$
$\displaystyle u = \frac{1}{x}$
$\displaystyle du = -\frac{1}{x^2} \, dx$
$\displaystyle -\int \sec^2\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) \, dx$
sub ...
$\displaystyle -\int \sec^2{u} \, du$
do the integration.
Here's exactly what you do for the first one so hopefully you can get the method down.
$\displaystyle \int\cos^3\theta\sin\theta d\theta$. So if you set $\displaystyle u=\cos\theta, \frac{du}{d\theta}=-\sin\theta$. So move over the $\displaystyle d\theta$, $\displaystyle du=-\sin\theta d\theta$. Solve for $\displaystyle d\theta$, $\displaystyle d\theta=-\frac{du}{\sin\theta}$. Substituting this in,
$\displaystyle \int\cos^3\theta\sin\theta d\theta=\int u^3\frac{-\sin\theta du}{\sin\theta}=-\int u^3 du=-\frac{1}{4}u^4 + C = -\frac{1}{4}\cos^4\theta + C.$
This follows after you remember what we defined u as.
