$I=\int\limits_{0}^{\frac{\Pi }{4}}\frac{dx}{\cos^{5} x\times\sin^{3} x}$
$I=\int\limits_{0}^{\frac{\Pi }{2}}\frac{\cos^{5} x}{\sqrt[3]{\sin^{2} x}}dx$
$I=\int\limits_{0}^{\Pi }\left ( \cos^{100} 2x\right )\left ( \sin^{7} 2x \right )dx$
$I=\int\limits_{0}^{\frac{\Pi }{2}}\sin^{8} 3xdx$
$I=\int\limits_{0}^{\frac{\Pi }{2}}\left ( 3+\cos x \right )^{5}dx$
$I=\int\limits_{0}^{\frac{\Pi }{2}}\left ( \cos 5x \right )^{13}dx$
$I=\int\limits_{0}^{\frac{\Pi }{4}}\left ( \sin 7x \right )^{11}dx$
e lên chia thành nhiều câu hỏi đăng mục hỏi đáp nhé ! để tiện cho các Admin giải bài. –  Đức Vỹ 11-11-14 10:36 AM
2 câu đầu cận $0$ làm cho hàm dưới dấu tích phân không xác định, xem lại đề

Tôi làm 1 hay 2 câu mà tôi thích thôi nhé, mấy bài này dài gõ mỏi tay

$I_3=\int \cos^{100} 2x .\sin^7 2x dx=-\dfrac{1}{2}\int \cos^{100} 2x . \sin^6 2x \ .d(\cos 2x)$

$=-\dfrac{1}{2} \int t^{100} (1-t^2)^3 dt$ Phá tung toét ra là xong thôi

$I_6=\int \cos^{13} 5x \ dx=\dfrac{1}{5}\int \cos^{12} 5x \ d(\sin 5x)=\dfrac{1}{5}\int (1-t^2)^6 dt$ Lại phá ra mà làm

Mấy câu kia cũng vậy cả

Bạn cần đăng nhập để có thể gửi đáp án

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