Tích Phân Của (3sinx+4cosx)/(4sinx+3cosx)dx Cân Từ Pi/2 đến 0 Tính ...

Đáp án:

\( \dfrac{{12\pi }}{{25}} + \dfrac{7}{{25}}\ln \dfrac{4}{3}\).

Giải thích các bước giải:

\(I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{3\sin x + 4\cos x}}{{4\sin x + 3\cos x}}dx} \)

Phân tích

\(\begin{array}{l}3\sin x + 4\cos x = A\left( {4\sin x + 3\cos x} \right) + B\left( {4\cos x - 3\sin x} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4A - 3B} \right)\sin x + \left( {3A + 4B} \right)\cos x\\ \Rightarrow \left\{ \begin{array}{l}4A - 3B = 3\\3A + 4B = 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}A = \dfrac{{24}}{{25}}\\B = \dfrac{7}{{25}}\end{array} \right.\\ \Rightarrow 3\sin x + 4\cos x = \dfrac{{24}}{{25}}\left( {4\sin x + 3\cos x} \right) + \dfrac{7}{{25}}\left( {4\cos x - 3\sin x} \right)\end{array}\)

\(\begin{array}{l} \Rightarrow I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{3\sin x + 4\cos x}}{{4\sin x + 3\cos x}}dx} \\I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\dfrac{{24}}{{25}}\left( {4\sin x + 3\cos x} \right) + \dfrac{7}{{25}}\left( {4\cos x - 3\sin x} \right)}}{{4\sin x + 3\cos x}}dx} \\\,\,\,\, = \dfrac{{24}}{{25}}\int\limits_0^{\dfrac{\pi }{2}} {dx} + \dfrac{7}{{25}}\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{4\cos x - 3\sin x}}{{4\sin x + 3\cos x}}dx} \\\,\,\,\, = \left. {\dfrac{{24}}{{25}}x} \right|_0^{\dfrac{\pi }{2}} + \dfrac{7}{{25}}{I_1}\\\,\,\,\, = \dfrac{{24}}{{25}}.\dfrac{\pi }{2} + \dfrac{7}{{25}}{I_1}\\\,\,\,\, = \dfrac{{12\pi }}{{25}} + \dfrac{7}{{25}}{I_1}\end{array}\)

Đặt \(t = 4\sin x + 3\cos x \Rightarrow dt = \left( {4\cos x - 3\sin x} \right)dx\)

Đổi cận: \(x = 0 \Rightarrow t = 3;\,\,x = \dfrac{\pi }{2} \Rightarrow t = 4\)

\( \Rightarrow {I_1} = \int\limits_3^4 {\dfrac{{dt}}{t}} = \left. {\ln \left| t \right|} \right|_3^4 = \ln 4 - \ln 3 = \ln \dfrac{4}{3}\)

Vậy \(I = \dfrac{{12\pi }}{{25}} + \dfrac{7}{{25}}\ln \dfrac{4}{3}\).