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Question Number 35058 by math khazana by abdo last updated on 14/May/18

$$find{\int }_0^{\frac{\pi }{4}}\frac{dt}{{(1+{cos}^{2}t)}^3}$$

Commented by abdo mathsup 649 cc last updated on 18/May/18

$$letputI={\int }_0^{\frac{\pi }{4}}\frac{dt}{{(1+{cos}^{2}t)}^3}$$$$I={\int }_0^{\frac{\pi }{4}}\frac{dt}{{(1+\frac{1+cos\left(2t\right)}{2})}^3}$$$$=8{\int }_0^{\frac{\pi }{4}}\frac{dt}{{(3+cos\left(2t\right))}^3}{=}_{2t=x}8{\int }_0^{\frac{\pi }{2}}\frac{1}{{(3+cosx)}^3}\frac{dx}{2}$$$$=4{\int }_0^{\frac{\pi }{2}}\frac{dx}{{(3+cosx)}^3}.changementtan\left(\frac{x}{2}\right)=t$$$$giveI=4{\int }_0^1\frac{1}{{(3+\frac{1-t^2}{1+t^2})}^3}\frac{2dt}{1+t^2}$$$$I=8{\int }_0^1\frac{{(1+t^2)}^2}{{(3+3t^2+1-t^2)}^3}dt$$$$I=8{\int }_0^1\frac{{(1+t^2)}^2}{{(4+2t^2)}^3}dt={\int }_0^1\frac{{(1+t^2)}^2}{{(2+t^2)}^3}dt$$$$I={\int }_0^1\frac{t^4+2t^2+1}{8+3.2^2{t}^2+3.2.t^4+t^6}dt$$$$={\int }_0^1\frac{t^4+2t^2+1}{t^6+6t^4+12t^2+8}dtletdecompose$$$$F\left(t\right)=\frac{t^4+2t^2+1}{t^6+6t^4+12t^2+8}.....becontinued...$$

Answered by MJS last updated on 16/May/18

$$\int \frac{dt}{{(1+{\cos }^2\left(t\right))}^3}=$$$$[\cos \left(t\right)=\frac{1}{\sec \left(t\right)};{\sec }^2\left(t\right)={\tan }^2\left(t\right)+1]$$$$=\int \frac{{\sec }^2\left(t\right){({\tan }^2\left(t\right)+1)}^2}{{({\tan }^2\left(t\right)+2)}^3}dt=$$$$[u=\tan \left(t\right)\to dt=\frac{du}{{\sec }^2\left(t\right)}]$$$$=\int \frac{{(u^2+1)}^2}{{(u^2+2)}^3}du=\int \frac{du}{u^2+2}-2\int \frac{du}{{(u^2+2)}^2}+\int \frac{du}{{(u^2+2)}^3}=$$$$$$$$\int \frac{du}{u^2+2}=$$$$[v=\frac{\sqrt{2}}{2}u\to du=dv\sqrt{2}]$$$$=\frac{\sqrt{2}}{2}\int \frac{dv}{v^2+1}=\frac{\sqrt{2}}{2}\arctan \left(v\right)=\frac{\sqrt{2}}{2}\arctan \left(\frac{\sqrt{2}}{2}u\right)$$$$$$$$-2\int \frac{du}{{(u^2+2)}^2}=$$$$[\int \frac{du}{{({au}^2+b)}^n}=\frac{u}{2b(n-1){({au}^2+b)}^{n-1}}+\frac{2n-3}{2b(n-1)}\int \frac{du}{{({au}^2+b)}^{n-1}}]$$$$=-2(\frac{u}{4(u^2+2)}+\frac{1}{4}\int \frac{du}{u^2+2})=$$$$=-\frac{u}{2(u^2+2)}-\frac{\sqrt{2}}{4}\arctan \left(\frac{\sqrt{2}}{2}u\right)$$$$$$$$\int \frac{du}{{(u^2+2)}^3}=\frac{u}{8{(u^2+2)}^2}+\frac{3}{8}\int \frac{du}{{(u^2+2)}^2}=$$$$=\frac{u}{8{(u^2+2)}^2}+\frac{3}{8}(\frac{u}{4(u^2+2)}+\frac{\sqrt{2}}{8}\arctan \left(\frac{\sqrt{2}}{2}u\right))=$$$$=\frac{u}{8{(u^2+2)}^2}+\frac{3u}{32(u^2+2)}+\frac{3\sqrt{2}}{64}\arctan \left(\frac{\sqrt{2}}{2}u\right)$$$$$$$$=\frac{19\sqrt{2}}{64}\arctan \left(\frac{\sqrt{2}}{2}u\right)-\frac{13u}{32(u^2+2)}+\frac{u}{8{(u^2+2)}^2}=$$$$=\frac{19\sqrt{2}}{64}\arctan \left(\frac{\sqrt{2}}{2}\tan \left(t\right)\right)-\frac{\tan \left(t\right)({13\tan }^2\left(t\right)+22)}{32{({\tan }^2\left(t\right)+2)}^2}+C$$$$$$$$\underset{0}{\stackrel{\frac{\pi }{4}}{\int }}\frac{dt}{{(1+{\cos }^2\left(t\right))}^3}=\frac{19\sqrt{2}}{64}\arctan \left(\frac{\sqrt{2}}{2}\right)-\frac{35}{288}$$

Commented by abdo mathsup 649 cc last updated on 15/May/18

$$thankyousirMjsforthishardwork...$$

Commented by MJS last updated on 16/May/18

$${\mathrm{you}}^{\prime }\mathrm{re}\mathrm{welcome},\mathrm{it}\mathrm{keeps}\mathrm{my}\mathrm{brains}\mathrm{young}$$

Commented by abdo mathsup 649 cc last updated on 16/May/18

$$areyouadoctororenjenersirMjs...$$

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