eseper je napisao:1.
[dispmath]\int\frac{\sin^2 x}{e^x}\mathrm dx[/dispmath]
Ovo sam išao rješvati preko parcijalne,
[inlmath]u=\frac{1}{e^x}\\
\mathrm du=-\frac{1}{e^x}\mathrm dx\\
\mathrm dv=\sin^2 x\mathrm dx\\
v=\int\sin^2 x\mathrm dx=\frac{x}{2}-\frac{1}{4}\sin(2x)+C[/inlmath]
itd. ali ovo definitivno nije dobar način jer se zadatak jako zakomplicira i nema mu kraja, iako se vjerovatno može rješiti ovako... Postoji li možda neki trik kojim bi ovo pojednostavnili, a da ga nisam uočio?
[inlmath]1.[/inlmath] način:
[dispmath]\int\frac{\sin^2 x}{e^x}\mathrm dx=[/dispmath]
[inlmath]u=\sin^2 x\\
\mathrm du=2\sin x\cos x\mathrm dx=\sin 2x\mathrm dx\\
\mathrm dv=e^{-x}\mathrm dx\\
v=-e^{-x}[/inlmath]
[dispmath]=-e^{-x}\sin^2 x+\int e^{-x}\sin 2x\mathrm dx\quad\left(1\right)[/dispmath][dispmath]\int e^{-x}\sin 2x\mathrm dx=[/dispmath]
[inlmath]u=\sin 2x\\
\mathrm du=2\cos 2x\mathrm dx\\
\mathrm dv=e^{-x}\mathrm dx\\
v=-e^{-x}[/inlmath]
[dispmath]=-e^{-x}\sin 2x+2\int e^{-x}\cos 2x\mathrm dx=[/dispmath]
[inlmath]u=\cos 2x\\
\mathrm du=-2\sin 2x\mathrm dx\\
\mathrm dv=e^{-x}\mathrm dx\\
v=-e^{-x}[/inlmath]
[dispmath]=-e^{-x}\sin 2x+2\left(-e^{-x}\cos 2x-2\int e^{-x}\sin 2x\mathrm dx\right)=-e^{-x}\sin 2x-2e^{-x}\cos 2x-4\int e^{-x}\sin 2x\mathrm dx[/dispmath][dispmath]\Rightarrow\quad 5\int e^{-x}\sin 2x\mathrm dx=-e^{-x}\sin 2x-2e^{-x}\cos 2x+c[/dispmath][dispmath]\int e^{-x}\sin 2x\mathrm dx=-\frac{1}{5}e^{-x}\sin 2x-\frac{2}{5}e^{-x}\cos 2x+c[/dispmath]
pa to vratimo u [inlmath]\left(1\right)[/inlmath]
[dispmath]-e^{-x}\sin^2 x-\frac{1}{5}e^{-x}\sin 2x-\frac{2}{5}e^{-x}\cos 2x+c=-e^{-x}\left(\sin^2 x+\frac{1}{5}\sin 2x+\frac{2}{5}\cos 2x\right)+c=[/dispmath][dispmath]=-e^{-x}\left(\sin^2 x+\frac{1}{5}\sin 2x+\frac{2}{5}\cos^2 x-\frac{2}{5}\sin^2 x\right)+c=[/dispmath][dispmath]=-e^{-x}\left(\sin^2 x+\frac{1}{5}\sin 2x+\frac{2}{5}-\frac{2}{5}\sin^2 x-\frac{2}{5}\sin^2 x\right)+c=[/dispmath][dispmath]=-e^{-x}\left(\frac{1}{5}\sin^2 x+\frac{1}{5}\sin 2x+\frac{2}{5}\right)+c=-\frac{e^{-x}}{5}\left(\sin^2 x+\sin 2x+2\right)+c[/dispmath]
[inlmath]2.[/inlmath] način:
[dispmath]\int\frac{\sin^2 x}{e^x}\mathrm dx=\int e^{-x}\frac{1-\cos 2x}{2}\mathrm dx=-\frac{1}{2}e^{-x}-\frac{1}{2}\int e^{-x}\cos 2x\mathrm dx\quad\left(1\right)[/dispmath][dispmath]\int e^{-x}\cos 2x\mathrm dx=[/dispmath]
[inlmath]u=\cos 2x\\
\mathrm du=-2\sin 2x\mathrm dx\\
\mathrm dv=e^{-x}\mathrm dx\\
v=-e^{-x}[/inlmath]
[dispmath]=-e^{-x}\cos 2x-2\int e^{-x}\sin 2x\mathrm dx=[/dispmath]
[inlmath]u=\sin 2x\\
\mathrm du=2\cos 2x\mathrm dx\\
\mathrm dv=e^{-x}\mathrm dx\\
v=-e^{-x}[/inlmath]
[dispmath]=-e^{-x}\cos 2x-2\left(-e^{-x}\sin 2x+2\int e^{-x}\cos 2x\mathrm dx\right)=-e^{-x}\cos 2x+2e^{-x}\sin 2x-4\int e^{-x}\cos 2x\mathrm dx=[/dispmath][dispmath]\Rightarrow\quad 5\int e^{-x}\cos 2x\mathrm dx=-e^{-x}\cos 2x+2e^{-x}\sin 2x+c[/dispmath][dispmath]\int e^{-x}\cos 2x\mathrm dx=-\frac{1}{5}e^{-x}\cos 2x+\frac{2}{5}e^{-x}\sin 2x+c[/dispmath]
pa to vratimo u [inlmath]\left(1\right)[/inlmath]
[dispmath]-\frac{1}{2}e^{-x}-\frac{1}{2}\left(-\frac{1}{5}e^{-x}\cos 2x+\frac{2}{5}e^{-x}\sin 2x\right)+c=-\frac{1}{2}e^{-x}+\frac{1}{10}e^{-x}\cos 2x-\frac{1}{5}e^{-x}\sin 2x+c=[/dispmath][dispmath]=-\frac{1}{10}e^{-x}\left(5-\cos 2x+2\sin 2x\right)+c=-\frac{1}{10}e^{-x}\left(5-\cos^2 x+\sin^2 x+2\sin 2x\right)+c=[/dispmath][dispmath]=-\frac{1}{10}e^{-x}\left(5-1+\sin^2 x+\sin^2 x+2\sin 2x\right)+c=-\frac{1}{10}e^{-x}\left(4+2\sin^2 x+2\sin 2x\right)+c=[/dispmath][dispmath]=-\frac{1}{5}e^{-x}\left(2+\sin^2 x+\sin 2x\right)+c[/dispmath]
A može i na način na koji si započeo:
[dispmath]\frac{xe^{-x}}{2}-\frac{1}{4}e^{-x}\sin 2x+\int\left(\frac{xe^{-x}}{2}-\frac{1}{4}e^{-x}\sin 2x\right)\mathrm dx=[/dispmath][dispmath]=\frac{xe^{-x}}{2}-\frac{1}{4}e^{-x}\sin 2x+\frac{1}{2}\int xe^{-x}\mathrm dx-\frac{1}{4}\int e^{-x}\sin 2x\mathrm dx\quad\left(1\right)[/dispmath][dispmath]\int xe^{-x}\mathrm dx=[/dispmath]
[inlmath]u=x\\
\mathrm du=\mathrm dx\\
\mathrm dv=e^{-x}\mathrm dx\\
v=-e^{-x}[/inlmath]
[dispmath]=-xe^{-x}+\int e^{-x}\mathrm dx=-xe^{-x}-e^{-x}+c\quad\left(2\right)[/dispmath]
U [inlmath]1.[/inlmath] načinu već je pokazano kako se izračunava [inlmath]\int e^{-x}\sin 2x\mathrm dx[/inlmath], pri čemu se dobije
[dispmath]\int e^{-x}\sin 2x\mathrm dx=-\frac{1}{5}e^{-x}\sin 2x-\frac{2}{5}e^{-x}\cos 2x+c\quad\left(3\right)[/dispmath]
Uvrstimo [inlmath]\left(2\right)[/inlmath] i [inlmath]\left(3\right)[/inlmath] u [inlmath]\left(1\right)[/inlmath]:
[dispmath]\frac{xe^{-x}}{2}-\frac{1}{4}e^{-x}\sin 2x+\frac{1}{2}\left(-xe^{-x}-e^{-x}\right)-\frac{1}{4}\left(-\frac{1}{5}e^{-x}\sin 2x-\frac{2}{5}e^{-x}\cos 2x\right)+c=[/dispmath][dispmath]=\cancel{\frac{xe^{-x}}{2}}-\frac{1}{4}e^{-x}\sin 2x-\cancel{\frac{1}{2}xe^{-x}}-\frac{1}{2}e^{-x}+\frac{1}{20}e^{-x}\sin 2x+\frac{1}{10}e^{-x}\cos 2x+c=[/dispmath][dispmath]=\frac{e^{-x}}{20}\left(-5\sin 2x-10+\sin 2x+2\cos 2x\right)+c=\frac{e^{-x}}{20}\left(-4\sin 2x-10+2\cos^2 x-2\sin^2 x\right)+c=[/dispmath][dispmath]=\frac{e^{-x}}{20}\left(-4\sin 2x-10+2-2\sin^2 x-2\sin^2 x\right)+c=\frac{e^{-x}}{20}\left(-4\sin 2x-8-4\sin^2 x\right)+c=[/dispmath][dispmath]=-\frac{e^{-x}}{5}\left(\sin 2x+2+\sin^2 x\right)+c[/dispmath]