Tính tích phân $I=\int\limits_{0}^{\frac{\pi}{4} }\frac{\sin x+\cos x}{3+\sin 2x} dx $
ko biết làm dạng này –  congiola_ktqd 12-11-12 09:12 PM
 $I=\int\limits_{0}^{\frac{\pi}{4} }\frac{\sin x+\cos x}{3+\sin 2x} dx =\frac{1}{4}\int\limits_{0}^{\frac{\pi}{4} }(\sin x+\cos x)\frac{2-\sin x+\cos x+2+\sin x-\cos x}{4-(\sin x- \cos x)^2} dx$
$=\frac{1}{4}\int\limits_{0}^{\frac{\pi}{4} }\frac{\sin x+\cos x}{\sin x-\cos x+2} dx+\frac{1}{4}\int\limits_{0}^{\frac{\pi}{4} }\frac{\sin x+\cos x}{-\sin x+\cos x+2} dx$
 $=\frac{1}{4}\ln\left| {\sin x-\cos x+2} \right|_{0}^{\frac{\pi}{4} }-\frac{1}{4}\ln\left| {-\sin x+\cos x+2} \right|_{0}^{\frac{\pi}{4} }$
 $I=\frac{1}{4}\ln3$
dạng này vừa học xong –  kellyhoang297 12-11-12 08:50 PM
Hãy ấn chữ V dưới đáp án để chấp nhận nếu như bạn thấy lời giải này chính xác, và nút mũi tên màu xanh để vote up nhé. Thanks! –  Trần Nhật Tân 11-11-12 10:34 PM
Ta có:
$I=\int\limits_0^{\frac{\pi}{4}}\frac{\sin x+\cos x}{3+\sin2x}dx$
    $=\frac{1}{4}\int\limits_0^{\frac{\pi}{4}}\left(\frac{\sin x+\cos x}{\sin x-\cos x+2}+\frac{\sin x+\cos x}{\cos x-\sin x+2}\right)dx$
    $=\frac{1}{4}\int\limits_0^{\frac{\pi}{4}}\left(\frac{d(\sin x-\cos x+2)}{\sin x-\cos x+2}-\frac{d(\cos x-\sin x+2)}{\cos x-\sin x+2}\right)$
    $=\frac{1}{4}\ln\frac{\sin x-\cos x+2}{\cos x-\sin x+2}\left|\begin{array}{l}\frac{\pi}{4}\\0\end{array}\right.=\frac{\ln3}{4}$

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