Ngô Quốc Anh

September 29, 2007

BT tính tích phân của hàm 1 biến

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 3:32

Tính \int^{\pi}_{-\pi} \frac{sin3x}{(1+2^{x})sinx} \,dx .

Lời giải: You can use

\boxed {\ \int_{ - a}^af(x)g(x)\ \mathrm {dx} = \int_0^a\left[f(x) + f( - x)\right]g(x)\ \mathrm {dx}\ },

where g is even. Therefore, for the functions f(x) = \frac {1}{1 + 2^x} and g(x) = \frac {\sin 3x}{\sin x} obtain :  and \int^{\pi}_{ - \pi} \frac {1}{1+2^x}\cdot\frac {\sin 3x}{\sin x}\ \mathrm {dx} = \int^{\pi}_0 \frac {\sin 3x}{\sin x}\ \mathrm {dx} a.s.o.  Here is another problem from Japanese university entrance exam. Keio University entrance exam/Medical 1974. Evaluate \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\sin ^ 2 x}{1+e^{-x}}\ dx.


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