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Calculeze

\( \displaystyle \left|\frac {\int_0^{\frac {\pi}{2}} (x\cos x + 1)e^{\sin x}\ dx}{\int_0^{\frac {\pi}{2}} (x\sin x - 1)e^{\cos x}\ dx}\right|. \)

You can see the solution here.

Fie \( I_n = \int_1^{1 + \frac {1}{n}}\{[(x + 1)\ln x + 1]e^{x(e^{x}\ln x + 1)} + n\}dx \ (n = 1,2,\cdots). \)

Calculati \( \lim_{n\to\infty}I_n^{n}. \)

Kunihiko Chikaya

Sa se calculeze

\( \int_0^1 (1+2008x^{2008})e^{x^{2008}}dx. \)

Kunihiko Chikaya

\( \int_0^{\pi} e^x\sin x\cos x\sin 2x\cos 2x dx \)

\( =\frac{1}{4}\int_0^{\pi} e^x\sin 2x\sin 4x dx \)

\( =\frac{1}{8}\int_0^{\pi} e^x(\cos 2x-\cos 6x)dx \)

\( =\frac{1}{8}\left[\frac{1}{5}e^x(\cos 2x+2\sin 2x)-\frac{1}{37}e^x(\cos 6x+6\sin 6x)\right]_0^{\pi} \)

\( =\frac{4}{185}(e^{\pi}-1) \)

I_0=\int \frac{1}{\sqrt{x^2+1}}\ dx=\ln |x+\sqrt{x^2+1}|+C. I_1=\int \frac{x}{\sqrt{x^2+1}}\ dx=\frac{1}{2}\int \frac{1}{\sqrt{x^2+1}}d(x^2+1)=\sqrt{x^2+1}+C. I_2=\int \frac{x^2}{\sqrt{x^2+1}}\ dx=\int \frac{(x^2+1)-1}{\sqrt{x^2+1}}\ dx =\int \left(\sqrt{x^2+1}-\frac{1}{\sqrt{x^2+1}}\right)\ dx =\i...