od Daniel » Ponedeljak, 09. Jun 2014, 21:31
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...
I do not fear death. I had been dead for billions and billions of years before I was born, and had not suffered the slightest inconvenience from it. – Mark Twain
, ja sam pokusao ovako ali ne mogu dalje 
