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[wieczyk]

\(\displaystyle{ \int e^x \sin^2 x dx}\)

[soku11]

\(\displaystyle{ =\int e^x \frac{1-cos2x}{2}dx=
\frac{1}{2}\int e^x (1-cos2x)dx=
\frac{1}{2}\int e^xdx-\frac{1}{2}\int e^xcos2xdx=
\frac{1}{2}e^x-\frac{1}{2}\int e^x cos2xdx\\
\mathcal{I}=\int e^xcos2xdx\\
u=e^x\ \ dv=cos2xdx\\
du=e^xdx\ \ v=\frac{1}{2}sin2x\\
\mathcal{I}=\frac{1}{2}e^xsin2x-\frac{1}{2} t e^xsin2xdx\\
u=e^x\ \ dv=sin2xdx\\
du=e^xdx\ \ v=-\frac{1}{2}cos2x\\
\mathcal{I}=\frac{1}{2}e^xsin2x-\frac{1}{2} (-\frac{1}{2} e^xcos2x+\frac{1}{2}\int e^xcos2xdx)=
\frac{1}{2}e^xsin2x+\frac{1}{4}e^xcos2x-\frac{1}{4}\int e^xcos2xdx=
\frac{1}{2}e^xsin2x+\frac{1}{4}e^xcos2x-\frac{1}{4}\mathcal{I}\\
\frac{5}{4}\mathcal{I}=\frac{1}{2}e^xsin2x+\frac{1}{4}e^xcos2x\\
\mathcal{I}=\frac{2}{5}e^xsin2x+\frac{1}{5}e^xcos2x+C}\)

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