Tính tích phân: $$\int\limits_{0}^{1}\left(\frac{\sin x.\cos x}{\sqrt{a^2\sin ^{2}x+b^2\cos x} }\right)$$
Để t thử xem làm được không

$I=\int \dfrac{\sin x \cos x}{\sqrt{a^2 (1-\cos^2 x) +b^2 \cos x}}dx=\int \dfrac{\sin x \cos x}{\sqrt{-a^2\cos^2 x +b^2 \cos x +a^2}}dx$

Đặt $\cos x=t \Rightarrow \sin x dx =-dt$

$I=-\int \dfrac{t}{\sqrt{-a^2 t^2 +b^2 t +a^2}}dt= \dfrac{1}{2a}\int \dfrac{-2at+b^2}{\sqrt{-a^2 t^2 +b^2 t +a^2}}dt-\dfrac{b^2}{2a} \int \dfrac{1}{\sqrt{-a^2 t^2 +b^2 t +a^2}}dt$

$= \dfrac{1}{2a}\int \dfrac{d(-a^2 t^2 +b^2 t +a^2)}{\sqrt{-a^2 t^2 +b^2 t +a^2}}dt-\dfrac{b^2}{2a} \int \dfrac{1}{\sqrt{-a^2 t^2 +b^2 t +a^2}}dt$

Toàn cái có dạng cả rồi nhé, tự làm đi, dài quá t không gõ hết đâu
uhm.dù sao cũng cảm ơn cậu nhiều :) –  nguyenthosp2 04-12-13 09:30 AM
:) hiểu cách làm là ok, tại bài lắm chữ quá lằng nhằng mà t thì k nháp :) –  Dép Lê Con Nhà Quê 04-12-13 08:58 AM
cách làm đúng rồi nhưng có chỗ nhầm.nhưng ko sao cảm ơn bạn nhiu nhiu nhé !!!!!!!!!!!!!! :) –  nguyenthosp2 04-12-13 08:54 AM

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