1. ## integration problem no.11

integration problem no.11

2. $\displaystyle \int\frac{1}{a^{2}cos^{2}{\theta}+b^{2}sin^{2}{\th eta}}d{\theta}$

Make the sub $\displaystyle u=tan{\theta}, \;\ tan^{-1}(u)={\theta}, \;\ d{\theta}=\frac{1}{u^{2}+1}du$

Then you get:

$\displaystyle \int\frac{1}{a^{2}+b^{2}u^{2}}du$

Now, let $\displaystyle w=tan^{-1}(\frac{ub}{a}), \;\ \frac{atan(w)}{b}=u, \;\ du=\frac{a}{bcos^{2}(w)}dw$

Making the subs whittles it down to:

$\displaystyle \int\frac{1}{ab}dw=\frac{w}{ab}$

Now make the resubs:

$\displaystyle \frac{tan^{-1}(\frac{ub}{a})}{ab}$

$\displaystyle \frac{tan^{-1}(\frac{btan({\theta})}{a})}{ab}$

3. Hello, afeasfaerw23231233!

A slight variation of Galactus' excellent solution . . .

$\displaystyle 22)\;\int^{\frac{\pi}{2}}_0\frac{d\theta}{a^2\cos^ 2\!\theta + b^2\sin^2\!\theta}$
Divide top and bottom by $\displaystyle \cos^2\!\theta\!:\;\;\int^{\frac{\pi}{2}}_0\frac{\ dfrac{d\theta}{\cos^2\theta}}{\dfrac{a^2\cos^2\the ta}{\cos^2\theta} + \dfrac{b^2\sin^2\theta}{\cos^2\theta}}$

. . and we have: .$\displaystyle \int^{\frac{\pi}{2}}_0\frac{\sec^2\theta\,d\theta} {a^2 + b^2\tan^2\theta}$

Let: $\displaystyle u \:=\:b\tan\theta\quad\Rightarrow\quad du \:=\:b\sec^2\theta\,d\theta$

Substitute: .$\displaystyle \frac{1}{b}\int\frac{du}{a^2+u^2} \;=\;\frac{1}{b}\cdot\frac{1}{a}\arctan\left(\frac {u}{a}\right) + C$

$\displaystyle \text{Back-substitute: }\;\frac{1}{ab}\arctan\left(\frac{b\tan\theta}{a}\ right) + C \;=\;\frac{1}{ab}\arctan\left(\dfrac{b}{a}\!\cdot\ !\tan\theta\right) + C$