Integeral Exercise 1/(sinx+2sec(x))^2 , asked in pretty mathematics group by Katrine linda

Exercise:

Integrate, $\int{\frac{dx}{{{\left( \sin x+2\sec x \right)}^{2}}}}$

Solution: we have $\frac{1}{{{\left( \sin x+2\sec x \right)}^{2}}}=\frac{{{\sec }^{2}}x}{{{\left( \sin x+2\sec x \right)}^{2}}{{\sec }^{2}}x}=\frac{{{\sec }^{2}}x}{4{{\sec }^{4}}x+4{{\sec }^{2}}x\tan x+{{\tan }^{2}}x}$

Let $u=\tan x\Leftrightarrow du={{\sec }^{2}}x\,dx$ But ${{\sec }^{2}}x={{\tan }^{2}}x+1\Leftrightarrow {{\sec }^{2}}x={{u}^{2}}+1$

So $\int{\frac{dx}{{{\left( \sin x+2\sec x \right)}^{2}}}=\int{\frac{{{\sec }^{2}}x}{4{{\sec }^{4}}x+4{{\sec }^{2}}x\tan x+{{\tan }^{2}}x}dx=\int{\frac{du}{4{{\sec }^{4}}x+4u{{\sec }^{2}}x+{{u}^{2}}}}}}$

$=\int{\frac{du}{4{{\left( {{u}^{2}}+1 \right)}^{2}}+4u\left( {{u}^{2}}+1 \right)+{{u}^{2}}}}=\int{\frac{du}{{{\left( 2{{u}^{2}}+u+2 \right)}^{2}}}}$

Now we need to take complete squaring for $2{{u}^{2}}+u+2$ to get ${{\left( \sqrt{2}\,u+\frac{1}{2\sqrt{2}} \right)}^{2}}+\frac{15}{8}$

So $\int{\frac{du}{{{\left( 2{{u}^{2}}+u+2 \right)}^{2}}}=\int{\frac{du}{{{\left( {{\left( \sqrt{2}u+\frac{1}{2\sqrt{2}} \right)}^{2}}+\frac{15}{8} \right)}^{2}}}}}$

Let $w=\sqrt{2}u+\frac{1}{2\sqrt{2}}\Leftrightarrow dw=\sqrt{2}\,du\Leftrightarrow \frac{dw}{\sqrt{2}}=du$

So $\int{\frac{du}{{{\left( {{\left( \sqrt{2}u+\frac{\sqrt{2}}{2} \right)}^{2}}+\frac{15}{8} \right)}^{2}}}}=\frac{1}{\sqrt{2}}\int{\frac{dw}{{{\left( {{w}^{2}}+\frac{15}{8} \right)}^{2}}}}$

Now let $w=\sqrt{\frac{15}{8}}\tan \theta \Leftrightarrow dw=\sqrt{\frac{15}{8}}{{\sec }^{2}}\theta \,d\theta $

Thus $\frac{1}{\sqrt{2}}\int{\frac{dw}{{{\left( {{w}^{2}}+\frac{15}{8} \right)}^{2}}}}=\frac{\sqrt{\frac{15}{8}}}{\sqrt{2}}\int{\frac{{{\sec }^{2}}\theta }{{{\left( \frac{15}{8}{{\tan }^{2}}\theta +\frac{15}{8} \right)}^{2}}}d\theta }=\frac{\sqrt{15}}{4}\int{\frac{{{\sec }^{2}}\theta }{{{\left( \frac{15}{8} \right)}^{2}}{{\left( {{\tan }^{2}}\theta +1 \right)}^{2}}}d\theta }$

$=\frac{\sqrt{15}}{4}\times \frac{64}{225}\int{\frac{1}{{{\sec }^{2}}\theta }d\theta =\frac{64\sqrt{15}}{900}\int{{{\cos }^{2}}\theta \,d\theta =\frac{64\sqrt{15}}{900}\frac{1}{2}\int{\left( 1+\cos 2\theta  \right)d\theta }}}$

$=\frac{64\sqrt{15}}{1800}\left( \theta +\frac{1}{2}\sin 2\theta  \right)+c$


but $\theta =\arctan \left( \sqrt{\frac{8}{15}}w \right)$ here we need to express $\sin 2\theta $ in terms of $\tan \theta $ .

We know that $\sin 2\theta =2\cos \theta \sin \theta \times 1=\frac{2{{\cos }^{2}}\theta \sin \theta }{\cos \theta }=

\frac{2\sin \theta }{\cos \theta }{{\cos }^{2}}\theta =2\tan \theta {{\cos }^{2}}\theta $

But $\sec \theta =\frac{1}{\cos \theta }\Leftrightarrow \cos \theta =\frac{1}{\sec \theta }\Leftrightarrow {{\cos }^{2}}\theta =\frac{1}{{{\sec }^{2}}\theta }=\frac{1}{{{\tan }^{2}}\theta +1}$ thus $\sin 2\theta =\frac{\tan \theta }{{{\tan }^{2}}\theta +1}$

Back word Substitution to get :

$\int{\frac{dx}{{{\left( \sin x+2\sec x \right)}^{2}}}=\frac{\left( \sin 2x+4 \right){{\sec }^{2}}x\left( 15\left( \cos 2x-15 \right)+8\sqrt{15}\left( \sin 2x+4 \right)\arctan \left( \frac{4\tan x+1}{\sqrt{15}} \right) \right)}{900{{\left( \sin x+2\sec x \right)}^{2}}}+c}$