nisam vidio temu sa ovim nazivom pa sam pomislio da bi bilo lijepo kad bi imalu i takvu temu
[dispmath]\begin{array}{lll}
628.\;\int\limits_0^4\frac{\ln\left(x+\sqrt{9+x^2}\right)}{\sqrt{9+x^2}}\mathrm dx & & 630.\;\int\limits_0^\frac{\pi}{2}3^{\cos^2 x}\sin 2x\mathrm dx \\
631.\;\int\limits_1^e\frac{\cos\left(\ln x\right)}{x}\mathrm dx\quad & 632.\;\int\limits_0^1\ln\left(x+1\right)\mathrm dx\qquad & 633.\;\int\limits_0^1 x\sinh x\mathrm dx
\end{array}[/dispmath]

Da, zaista bi bilo lepo, to je ono što je ovom forumu sve vreme falilo, ali sad imamo i to.

slavonija035 je napisao:[dispmath]628.\;\int\limits_0^4\frac{\ln\left(x+\sqrt{9+x^2}\right)}{\sqrt{9+x^2}}\mathrm dx[/dispmath]

Uvedemo smenu
[inlmath]x=3\sinh t\\
\mathrm dx=3\cosh t\mathrm dt[/inlmath]
i odredimo nove granice integraljenja:

[inlmath]\underline{x=0:}\\
3\sinh t=0\quad\Rightarrow\quad 3\frac{e^t-e^{-t}}{2}=0\quad\Rightarrow\quad e^t=e^{-t}\quad\Rightarrow\quad\left(e^t\right)^2=1\quad\Rightarrow\quad t=0[/inlmath]

[inlmath]\underline{x=4:}\\
3\sinh t=4\quad\Rightarrow\quad 3\frac{e^t-e^{-t}}{2}=4\quad\Rightarrow\quad 3e^t-3e^{-t}-8=0\quad\Rightarrow\\
\Rightarrow\quad 3\left(e^t\right)^2-8e^t-3=0\quad\Rightarrow\quad\left(e^t\right)_1=-\frac{1}{3},\;\left(e^t\right)_2=3[/inlmath]
[inlmath]\left(e^t\right)_1[/inlmath] odbacujemo, jer [inlmath]e^t[/inlmath] može imati samo pozitivne vrednosti, tako da ostaje [inlmath]e^t=3[/inlmath].
[inlmath]t=\ln 3[/inlmath]

Integral postaje:
[dispmath]\int\limits_0^{\ln 3}\frac{\ln\left(3\sinh t+\sqrt{9+9\sinh^2 t}\right)}{\sqrt{9+9\sinh^2 t}}\cdot 3\cosh t\mathrm dt=\int\limits_0^{\ln 3}\frac{\ln\left(3\sinh t+3\sqrt{1+\sinh^2 t}\right)}{3\sqrt{1+\sinh^2 t}}\cdot 3\cosh t\mathrm dt=[/dispmath][dispmath]=\int\limits_0^{\ln 3}\frac{\ln\left(3\sinh t+3\cosh t\right)}{3\cosh t}\cdot 3\cosh t\mathrm dt=\int\limits_0^{\ln 3}\ln\left(3\sinh t+3\cosh t\right)\mathrm dt=[/dispmath][dispmath]=\int\limits_0^{\ln 3}\ln\left[3\left(\sinh t+\cosh t\right)\right]\mathrm dt=\int\limits_0^{\ln 3}\left[\ln 3+\ln\left(\sinh t+\cosh t\right)\right]\mathrm dt=[/dispmath][dispmath]=\ln 3\int\limits_0^{\ln 3}\mathrm dt+\int\limits_0^{\ln 3}\ln\left(\sinh t+\cosh t\right)\mathrm dt=\ln 3\cdot\left.t\right|_0^{\ln 3}+\int\limits_0^{\ln 3}\ln\left(\frac{e^t-e^{-t}}{2}+\frac{e^t+e^{-t}}{2}\right)\mathrm dt=[/dispmath][dispmath]=\ln^2 3+\int\limits_0^{\ln 3}\ln e^t\mathrm dt=\ln^2 3+\int\limits_0^{\ln 3}t\mathrm dt=\ln^2 3+\left.\frac{t^2}{2}\right|_0^{\ln 3}=\ln^2 3+\frac{\ln^2 3}{2}=\frac{3}{2}\ln^2 3[/dispmath]

slavonija035 je napisao:[dispmath]630.\;\int\limits_0^\frac{\pi}{2}3^{\cos^2 x}\sin 2x\mathrm dx[/dispmath]

[dispmath]\int\limits_0^\frac{\pi}{2}3^{\cos^2 x}\sin 2x\mathrm dx=\int\limits_0^\frac{\pi}{2}3^{\cos^2 x}\cdot 2\sin x\cos x\mathrm dx=-\int\limits_0^\frac{\pi}{2}3^{\cos^2 x}\cdot 2\cos x\mathrm d\left(\cos x\right)=[/dispmath][dispmath]=-\int\limits_0^\frac{\pi}{2}3^{\cos^2 x}\mathrm d\left(\cos^2 x\right)=-\frac{1}{\ln 3}\left.3^{\cos^2 x}\right|_0^\frac{\pi}{2}=-\frac{1}{\ln 3}\left(3^{\cos^2 \frac{\pi}{2}}-3^{\cos^2 0}\right)=[/dispmath][dispmath]=-\frac{1}{\ln 3}\left(3^0-3^1\right)=-\frac{1}{\ln 3}\left(1-3\right)=-\frac{1}{\ln 3}\left(3^0-3^1\right)=-\frac{1}{\ln 3}\left(1-3\right)=\frac{2}{\ln 3}[/dispmath]

slavonija035 je napisao:[dispmath]631.\;\int\limits_1^e\frac{\cos\left(\ln x\right)}{x}\mathrm dx[/dispmath]

[dispmath]\int\limits_1^e\frac{\cos\left(\ln x\right)}{x}\mathrm dx=\int\limits_1^e\cos\left(\ln x\right)\mathrm d\left(\ln x\right)=\left.\sin\left(\ln x\right)\right|_1^e=[/dispmath][dispmath]=\sin\left(\ln e\right)-\sin\left(\ln 1\right)=\sin 1-\sin 0=\sin 1[/dispmath]

slavonija035 je napisao:[dispmath]632.\;\int\limits_0^1\ln\left(x+1\right)\mathrm dx[/dispmath]

[inlmath]u=\ln\left(x+1\right)\\
\mathrm du=\frac{\mathrm dx}{x+1}\\
\mathrm dv=\mathrm dx\\
v=x[/inlmath]
[dispmath]=\left.x\ln\left(x+1\right)\right|_0^1-\int\limits_0^1\frac{x}{x+1}\mathrm dx=\ln 2-\int\limits_0^1\frac{x+1-1}{x+1}\mathrm dx=\ln 2-\int\limits_0^1\mathrm dx+\int\limits_0^1\frac{\mathrm dx}{x+1}=[/dispmath][dispmath]=\ln 2-\left.x\right|_0^1+\left.\ln\left|x+1\right|\right|_0^1=\ln 2-1+\ln 2=2\ln 2-1[/dispmath]

slavonija035 je napisao:[dispmath]633.\;\int\limits_0^1 x\sinh x\mathrm dx[/dispmath]

[inlmath]u=x\\
\mathrm du=\mathrm dx\\
\mathrm dv=\sinh x\mathrm dx\\
v=\cosh x[/inlmath]
[dispmath]=\left.x\cosh x\right|_0^1-\int\limits_0^1\cosh x\mathrm dx=\cosh 1-\left.\sinh x\right|_0^1=\cosh 1-\sinh x 1=\frac{e^1+e^{-1}}{2}-\frac{e^1-e^{-1}}{2}=e^{-1}=\frac{1}{e}[/dispmath]