MATEMANIJA

Treba mi ideja za ovaj zadatak , ja sam pokusao ovako ali ne mogu dalje
[dispmath]\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^4x\sin^2x\mathrm dx=\begin{vmatrix}
\cos x=t\\ \sin x=\mathrm dt
\end{vmatrix}=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}t^4\mathrm dt\sin x=\cdots\;?[/dispmath]

Nisam baš uhodan za integralima, al' čini mi se da bi ovako trebalo:
[dispmath]\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\cos^4x\sin^2x\mathrm dx=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\cos^4x\left(1-\cos^2x\right)\mathrm dx=\\ \int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\left(\cos^4x-\cos^6x\right)\mathrm dx=\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\cos^4x\mathrm dx-\int\limits^\frac{\pi}{2}_{-\frac{\pi}{2}}\cos^6x\mathrm dx\\[3em][/dispmath]
E, sad, ne znam koja se smena uvodi i da l' uopšte treba smena, pošto bih ja to "na suvo" odradio.

Dalje mozes da predstavis [inlmath]\cos^2x=\frac{1}{\mathrm{tg}^2x+1}[/inlmath],uvedes smenu [inlmath]\mathrm{tg}\:x=t[/inlmath] i dalje resavas integral racionalne funkcije...

a kako da resim [inlmath]\cos^6[/inlmath]

Ja bih radio tako što bih i na [inlmath]\cos^4x[/inlmath] i na [inlmath]\sin^2x[/inlmath] primenio formulu za kosinus, odnosno sinus, polovine ugla:
[dispmath]\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^4x\sin^2x\mathrm dx=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{1+\cos 2x}{2}\right)^2\cdot\frac{1-\cos 2x}{2}\mathrm dx=\frac{1}{8}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(1+\cos 2x\right)\underbrace{\left(1+\cos 2x\right)\left(1-\cos 2x\right)}\mathrm dx=[/dispmath][dispmath]=\frac{1}{8}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(1+\cos 2x\right)\left(1-\cos^22x\right)\mathrm dx=\frac{1}{8}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(1+\cos 2x\right)\sin^22x\mathrm dx=[/dispmath][dispmath]=\frac{1}{8}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^22x\mathrm dx+\frac{1}{8}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^22x\cos 2x\mathrm dx=\frac{1}{8}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1-\cos 4x}{2}\mathrm dx+\frac{1}{16}\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^22x\mathrm d\left(\sin 2x\right)=\cdots[/dispmath]
Dalje, pretpostavljam, nije problem...

Ako zelis da radis na nacin koji je Stevan predlozio, mozes nastaviti tako sto iskoristis da je
[dispmath]\cos^4x=\frac{1}{8}\left(\cos 4x+4\cos 2x+3\right)[/dispmath][dispmath]\cos^6x=\frac{1}{32}\left(10+15\cos 2x+6\cos 4x+\cos 6x\right)[/dispmath]

Ako te buni kako je Milovan došao do ovih izraza – evo kako:
[dispmath]\cos^4x=\left(\cos^2x\right)^2=\left(\frac{1+\cos 2x}{2}\right)^2=\frac{1}{4}\left(1+2\cos 2x+\cos^22x\right)=[/dispmath][dispmath]=\frac{1}{4}\left(1+2\cos 2x+\frac{1+\cos 4x}{2}\right)=\frac{1}{8}\left(2+4\cos 2x+1+\cos 4x\right)=\frac{1}{8}\left(\cos 4x+4\cos 2x+3\right)[/dispmath]
[dispmath]\cos^6x=\left(\cos^2x\right)^3=\left(\frac{1+\cos 2x}{2}\right)^3=\frac{1}{8}\left(1+3\cos 2x+3\cos^22x+\cos^32x\right)=[/dispmath][dispmath]=\frac{1}{8}\left(1+3\cos 2x+3\cdot\frac{1+\cos 4x}{2}+\frac{1+\cos 4x}{2}\cdot\cos 2x\right)=[/dispmath][dispmath]=\frac{1}{16}\left(2+6\cos 2x+3+3\cos 4x+\cos 2x+\cos 4x\cos 2x\right)=[/dispmath][dispmath]=\frac{1}{16}\left[5+7\cos 2x+3\cos 4x+\frac{1}{2}\left(\cos 6x+\cos 2x\right)\right]=[/dispmath][dispmath]=\frac{1}{32}\left(10+14\cos 2x+6\cos 4x+\cos 6x+\cos 2x\right)=[/dispmath][dispmath]=\frac{1}{32}\left(10+15\cos 2x+6\cos 4x+\cos 6x\right)[/dispmath]