integration problem no.11

**afeasfaerw23231233** · November 22nd 2007, 12:58 AM

**galactus** · November 22nd 2007, 03:10 AM

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

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

Then you get:

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

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

Making the subs whittles it down to:

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

Now make the resubs:

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

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

**Soroban** · November 22nd 2007, 04:02 AM

Hello, afeasfaerw23231233!

A slight variation of Galactus' excellent solution . . .

$22)\;\int^{\frac{\pi}{2}}_0\frac{d\theta}{a^2\cos^ 2\!\theta + b^2\sin^2\!\theta}$

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

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

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

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

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

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