2 replies to this topic

[vantrung1001]

I= $\int_{0}^{\frac{\pi}{2}}\frac{sin2x}{1+cos^{2}x}dx$

 

J= $\int_{0}^{1}\frac{xdx}{x+\sqrt{x^2+1}}$

 

H = $\int_{\frac{\pi }{4}}^{\frac{\pi}{2}}\frac{cos^2x}{sin^4x}dx$

Edited by vantrung1001, 23-01-2014 - 20:49.

[vuvanquya1nct]

I= $\int_{0}^{\frac{\pi}{2}}\frac{sin2x}{1+cos^{2}x}dx$

 

J= $\int_{0}^{1}\frac{xdx}{x+\sqrt{x^2+1}}$

 

H = $\int_{\frac{\pi }{4}}^{\frac{\pi}{2}}\frac{cos^2x}{sin^4x}dx$

Bài 1 đặt $t=cosx$

$I=2\int_{0}^{1}\frac{tdt}{1+t^2}$=$\int_{0}^{1}\frac{d(1+t^2)}{1+t^2}$

Đến đây thay cận

Bài 2

Nhân lượng liên hợp

Edited by vuvanquya1nct, 23-01-2014 - 21:14.

[Enzan]

$H = \int_{\frac{\pi }{4}}^{\frac{\pi }{2}}\frac{cos^{2}x}{sin^{4}x}dx=\int_{\frac{\pi }{4}}^{\frac{\pi }{2}}\frac{cot^{2}x}{sin^{2}x}dx$

Đặt $t=cotx$ $\Rightarrow dt=\frac{-1}{sin^{2}x}dx$

 

$\Rightarrow$ $H =\int_{1}^{0}-t^{2}dt =\frac{1}{3}$

 

 

Edited by Enzan, 25-01-2014 - 23:25.