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được thưởng
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Đăng nhập hàng ngày 26/01/2014
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được thưởng
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Đăng nhập hàng ngày 23/01/2014
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sửa đổi
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bài tích phân
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$\Leftrightarrow \int\limits_{0}^{1}\frac{x^2+1-2x^2}{(1+x^2)^2}dx=\int\limits_{0}^{1}\frac{dx}{1+x^2}-2\int\limits_{0}^{1}\frac{x^2dx}{(1+x^2)^2}$$\Leftrightarrow\int\limits_{0}^{1}\frac{dx}{1+x^2}-2\int\limits_{0}^{1}\frac{x^2+1-1}{(1+x^2)^2}dx$$\Leftrightarrow2\int\limits_{0}^{1}\frac{dx}{(1+x^2)^2}+3\int\limits_{0}^{1}\frac{dx}{1+x^2}$Đặt $x=tanu$$\Leftrightarrow\begin{cases}dx=\frac{du}{cos^2u} \\x^2+1=\frac{1}{cos^2u}\end{cases}$$\Rightarrow 2\int\limits_{0}^{\frac{\pi }{4}}\frac{cos^4u.du}{cos^2u}-\int\limits_{0}^{\frac{\pi }{4}}du$$\Leftrightarrow2\int\limits_{0}^{\frac{\pi }{4}}\cos^2u.du$$\Leftrightarrow\int\limits_{0}^{\frac{\pi }{4}}cos2u.du=\frac{1}{2}sin2u|^{\frac{\pi}{4}}_{0}$
$\Leftrightarrow \int\limits_{0}^{1}\frac{x^2+1-2x^2}{(1+x^2)^2}dx=\int\limits_{0}^{1}\frac{dx}{1+x^2}-2\int\limits_{0}^{1}\frac{x^2dx}{(1+x^2)^2}$$\Leftrightarrow\int\limits_{0}^{1}\frac{dx}{1+x^2}-2\int\limits_{0}^{1}\frac{x^2+1-1}{(1+x^2)^2}dx$$\Leftrightarrow2\int\limits_{0}^{1}\frac{dx}{(1+x^2)^2}-\int\limits_{0}^{1}\frac{dx}{1+x^2}$Đặt $x=tanu$$\Leftrightarrow\begin{cases}dx=\frac{du}{cos^2u} \\x^2+1=\frac{1}{cos^2u}\end{cases}$$\Rightarrow 2\int\limits_{0}^{\frac{\pi }{4}}\frac{cos^4u.du}{cos^2u}-\int\limits_{0}^{\frac{\pi }{4}}du$$\Leftrightarrow2\int\limits_{0}^{\frac{\pi }{4}}\cos^2u.du$$\Leftrightarrow\int\limits_{0}^{\frac{\pi }{4}}cos2u.du=\frac{1}{2}sin2u|^{\frac{\pi}{4}}_{0}$
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sửa đổi
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bài tích phân
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$\Leftrightarrow \int\limits_{0}^{1}\frac{x^2+1-2x^2}{(1+x^2)^2}dx=\int\limits_{0}^{1}\frac{dx}{1+x^2}-2\int\limits_{0}^{1}\frac{x^2dx}{(1+x^2)^2}$$\Leftrightarrow\int\limits_{0}^{1}\frac{dx}{1+x^2}-2\int\limits_{0}^{1}\frac{x^2+1-1}{(1+x^2)^2}dx$$\Leftrightarrow2\int\limits_{0}^{1}\frac{dx}{(1+x^2)^2}+3\int\limits_{0}^{1}\frac{dx}{1+x^2}$Đặt $x=tanu$$\Leftrightarrow\begin{cases}dx=\frac{du}{cos^2u} \\x^2+1=\frac{1}{cos^2u}\end{cases}$$\Rightarrow 2\int\limits_{0}^{\frac{\pi }{4}}\frac{cos^4u.du}{cos^2u}-\int\limits_{0}^{\frac{\pi }{4}}du$$\Leftrightarrow2\int\limits_{0}^{\frac{\pi }{4}}\cos^2u.du+3\int\limits_{0}^{\frac{\pi }{4}}du$$\Leftrightarrow\int\limits_{0}^{\frac{\pi }{4}}cos2u.du+4\int\limits_{0}^{\frac{\pi }{4}}du=\frac{1}{2}sin2u|^{\frac{\pi}{4}}_{0}+4u|^{\frac{\pi}{4}}_{0} $
$\Leftrightarrow \int\limits_{0}^{1}\frac{x^2+1-2x^2}{(1+x^2)^2}dx=\int\limits_{0}^{1}\frac{dx}{1+x^2}-2\int\limits_{0}^{1}\frac{x^2dx}{(1+x^2)^2}$$\Leftrightarrow\int\limits_{0}^{1}\frac{dx}{1+x^2}-2\int\limits_{0}^{1}\frac{x^2+1-1}{(1+x^2)^2}dx$$\Leftrightarrow2\int\limits_{0}^{1}\frac{dx}{(1+x^2)^2}+3\int\limits_{0}^{1}\frac{dx}{1+x^2}$Đặt $x=tanu$$\Leftrightarrow\begin{cases}dx=\frac{du}{cos^2u} \\x^2+1=\frac{1}{cos^2u}\end{cases}$$\Rightarrow 2\int\limits_{0}^{\frac{\pi }{4}}\frac{cos^4u.du}{cos^2u}-\int\limits_{0}^{\frac{\pi }{4}}du$$\Leftrightarrow2\int\limits_{0}^{\frac{\pi }{4}}\cos^2u.du$$\Leftrightarrow\int\limits_{0}^{\frac{\pi }{4}}cos2u.du=\frac{1}{2}sin2u|^{\frac{\pi}{4}}_{0}$
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sửa đổi
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bài tích phân
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$\Leftrightarrow \int\limits_{0}^{1}\frac{x^2+1-2x^2}{(1+x^2)^2}dx=\int\limits_{0}^{1}\frac{dx}{1+x^2}-2\int\limits_{0}^{1}\frac{x^2dx}{(1+x^2)^2}$$\Leftrightarrow\int\limits_{0}^{1}\frac{dx}{1+x^2}-2\int\limits_{0}^{1}\frac{x^2+1-1}{(1+x^2)^2}dx$$\Leftrightarrow2\int\limits_{0}^{1}\frac{dx}{(1+x^2)^2}-\int\limits_{0}^{1}\frac{dx}{1+x^2}$Đặt $x=tanu$$\Leftrightarrow\begin{cases}dx=\frac{du}{cos^2u} \\x^2+1=\frac{1}{cos^2u}\end{cases}$$\Rightarrow 2\int\limits_{0}^{\frac{\pi }{4}}\frac{cos^4u.du}{cos^2u}-\int\limits_{0}^{\frac{\pi }{4}}du$$\Leftrightarrow2\int\limits_{0}^{\frac{\pi }{4}}\cos^2u.du-\int\limits_{0}^{\frac{\pi }{4}}du$$\Leftrightarrow\int\limits_{0}^{\frac{\pi }{4}}cos2u.du=\frac{1}{2}sin2u|^{\frac{\pi}{4}}_{0} $
$\Leftrightarrow \int\limits_{0}^{1}\frac{x^2+1-2x^2}{(1+x^2)^2}dx=\int\limits_{0}^{1}\frac{dx}{1+x^2}-2\int\limits_{0}^{1}\frac{x^2dx}{(1+x^2)^2}$$\Leftrightarrow\int\limits_{0}^{1}\frac{dx}{1+x^2}-2\int\limits_{0}^{1}\frac{x^2+1-1}{(1+x^2)^2}dx$$\Leftrightarrow2\int\limits_{0}^{1}\frac{dx}{(1+x^2)^2}+3\int\limits_{0}^{1}\frac{dx}{1+x^2}$Đặt $x=tanu$$\Leftrightarrow\begin{cases}dx=\frac{du}{cos^2u} \\x^2+1=\frac{1}{cos^2u}\end{cases}$$\Rightarrow 2\int\limits_{0}^{\frac{\pi }{4}}\frac{cos^4u.du}{cos^2u}-\int\limits_{0}^{\frac{\pi }{4}}du$$\Leftrightarrow2\int\limits_{0}^{\frac{\pi }{4}}\cos^2u.du+3\int\limits_{0}^{\frac{\pi }{4}}du$$\Leftrightarrow\int\limits_{0}^{\frac{\pi }{4}}cos2u.du+4\int\limits_{0}^{\frac{\pi }{4}}du=\frac{1}{2}sin2u|^{\frac{\pi}{4}}_{0}+4u|^{\frac{\pi}{4}}_{0} $
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giải đáp
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bài tích phân
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$\Leftrightarrow \int\limits_{0}^{1}\frac{x^2+1-2x^2}{(1+x^2)^2}dx=\int\limits_{0}^{1}\frac{dx}{1+x^2}-2\int\limits_{0}^{1}\frac{x^2dx}{(1+x^2)^2}$
$\Leftrightarrow\int\limits_{0}^{1}\frac{dx}{1+x^2}-2\int\limits_{0}^{1}\frac{x^2+1-1}{(1+x^2)^2}dx$
$\Leftrightarrow2\int\limits_{0}^{1}\frac{dx}{(1+x^2)^2}-\int\limits_{0}^{1}\frac{dx}{1+x^2}$
Đặt $x=tanu$
$\Leftrightarrow\begin{cases}dx=\frac{du}{cos^2u} \\x^2+1=\frac{1}{cos^2u}\end{cases}$
$\Rightarrow 2\int\limits_{0}^{\frac{\pi }{4}}\frac{cos^4u.du}{cos^2u}-\int\limits_{0}^{\frac{\pi }{4}}du$
$\Leftrightarrow2\int\limits_{0}^{\frac{\pi }{4}}\cos^2u.du$
$\Leftrightarrow\int\limits_{0}^{\frac{\pi }{4}}cos2u.du=\frac{1}{2}sin2u|^{\frac{\pi}
{4}}_{0}$
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bình luận
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Giải giúp mình với!!
nhưng cái đáp án x =....nó nhìn khác nhau mình không biết đáp án giải theo cách như bạn có giống với đáp án bộ không thôi. Vậy là giống nhau hả?
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được thưởng
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Đăng nhập hàng ngày 22/01/2014
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sửa đổi
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Tính tích phân
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$\Leftrightarrow \frac{1}{2}\int\limits_{0}^{\pi }e^xdx+\frac{1}{2}\int\limits_{0}^{\pi }cos2x.e^xdx$xét $J=\int\limits_{0}^{\pi }cos2x.e^xdx=\int\limits_{0}^{\pi }cos2x.d(e^x)$$\Leftrightarrow e^xcos2x-\int\limits_{0}^{\pi }e^xd(cos2x)$$\Leftrightarrow e^xcos2x+2\int\limits_{0}^{\pi }e^xsin2x.dx$$\Leftrightarrow e^xcos2x+2e^xsin2x-2\int\limits_{0}^{\pi }e^xd(sin2x)$$\Leftrightarrow e^xcos2x+2e^xsin2x+4\int\limits_{0}^{\pi }cos2x.e^xdx$Ta có: $J= e^xcos2x+2e^xsin2x+4\int\limits_{0}^{\pi }cos2x.e^xdx$$\Leftrightarrow\int\limits_{0}^{\pi }cos2x.e^xdx=e^xcos2x+2e^xsin2x+4\int\limits_{0}^{\pi }cos2x.e^xdx$$\Leftrightarrow\int\limits_{0}^{\pi }cos2x.e^xdx=-\frac{1}{5}(e^xcos2x+2e^xsin2x)$$\Rightarrow \int\limits_{0}^{\pi }cos^2x.e^xdx=\frac{1}{2}e^x|^{\pi }_0-\frac{1}{10}(e^xcos2x+2e^xsin2x)|^{\pi }_0$
$\Leftrightarrow \frac{1}{2}\int\limits_{0}^{\pi }e^xdx+\frac{1}{2}\int\limits_{0}^{\pi }cos2x.e^xdx$xét $J=\int\limits_{0}^{\pi }cos2x.e^xdx=\int\limits_{0}^{\pi }cos2x.d(e^x)$$\Leftrightarrow e^xcos2x-\int\limits_{0}^{\pi }e^xd(cos2x)$$\Leftrightarrow e^xcos2x+2\int\limits_{0}^{\pi }e^xsin2x.dx$$\Leftrightarrow e^xcos2x+2e^xsin2x-2\int\limits_{0}^{\pi }e^xd(sin2x)$$\Leftrightarrow e^xcos2x+2e^xsin2x+4\int\limits_{0}^{\pi }cos2x.e^xdx$Ta có: $J= e^xcos2x+2e^xsin2x+4\int\limits_{0}^{\pi }cos2x.e^xdx$$\Leftrightarrow\int\limits_{0}^{\pi }cos2x.e^xdx=e^xcos2x+2e^xsin2x+4\int\limits_{0}^{\pi }cos2x.e^xdx$$\Leftrightarrow\int\limits_{0}^{\pi }cos2x.e^xdx=-\frac{1}{3}(e^xcos2x+2e^xsin2x)$$\Rightarrow \int\limits_{0}^{\pi }cos^2x.e^xdx=\frac{1}{2}e^x|^{\pi }_0-\frac{1}{6}(e^xcos2x+2e^xsin2x)|^{\pi }_0$
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sửa đổi
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Tính tích phân
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$\Leftrightarrow \frac{1}{2}\int\limits_{0}^{\pi }e^xdx+\frac{1}{2}\int\limits_{0}^{\pi }cos2x.e^xdx$xét $J=\int\limits_{0}^{\pi }cos2x.e^xdx=\int\limits_{0}^{\pi }cos2x.d(e^x)$$\Leftrightarrow e^xcos2.-\int\limits_{0}^{\pi }e^xd(cos2x)$$\Leftrightarrow e^xcos2x+2\int\limits_{0}^{\pi }e^xsin2x.dx$$\Leftrightarrow e^xcos2x+2e^xsin2x-2\int\limits_{0}^{\pi }e^xd(sin2x)$$\Leftrightarrow e^xcos2x+2e^xsin2x+4\int\limits_{0}^{\pi }cos2x.e^xdx$Ta có: $J= e^xcos2x+2e^xsin2x+4\int\limits_{0}^{\pi }cos2x.e^xdx$$\Leftrightarrow\int\limits_{0}^{\pi }cos2x.e^xdx=e^xcos2x+2e^xsin2x+4\int\limits_{0}^{\pi }cos2x.e^xdx$$\Leftrightarrow\int\limits_{0}^{\pi }cos2x.e^xdx=-\frac{1}{5}(e^xcos2x+2e^xsin2x)$$\Rightarrow \int\limits_{0}^{\pi }cos^2x.e^xdx=\frac{1}{2}e^x|^{\pi }_0-\frac{1}{10}(e^xcos2x+2e^xsin2x)|^{\pi }_0$
$\Leftrightarrow \frac{1}{2}\int\limits_{0}^{\pi }e^xdx+\frac{1}{2}\int\limits_{0}^{\pi }cos2x.e^xdx$xét $J=\int\limits_{0}^{\pi }cos2x.e^xdx=\int\limits_{0}^{\pi }cos2x.d(e^x)$$\Leftrightarrow e^xcos2x-\int\limits_{0}^{\pi }e^xd(cos2x)$$\Leftrightarrow e^xcos2x+2\int\limits_{0}^{\pi }e^xsin2x.dx$$\Leftrightarrow e^xcos2x+2e^xsin2x-2\int\limits_{0}^{\pi }e^xd(sin2x)$$\Leftrightarrow e^xcos2x+2e^xsin2x+4\int\limits_{0}^{\pi }cos2x.e^xdx$Ta có: $J= e^xcos2x+2e^xsin2x+4\int\limits_{0}^{\pi }cos2x.e^xdx$$\Leftrightarrow\int\limits_{0}^{\pi }cos2x.e^xdx=e^xcos2x+2e^xsin2x+4\int\limits_{0}^{\pi }cos2x.e^xdx$$\Leftrightarrow\int\limits_{0}^{\pi }cos2x.e^xdx=-\frac{1}{5}(e^xcos2x+2e^xsin2x)$$\Rightarrow \int\limits_{0}^{\pi }cos^2x.e^xdx=\frac{1}{2}e^x|^{\pi }_0-\frac{1}{10}(e^xcos2x+2e^xsin2x)|^{\pi }_0$
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giải đáp
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Tính tích phân
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$\Leftrightarrow \frac{1}{2}\int\limits_{0}^{\pi }e^xdx+\frac{1}{2}\int\limits_{0}^{\pi }cos2x.e^xdx$
xét $J=\int\limits_{0}^{\pi }cos2x.e^xdx=\int\limits_{0}^{\pi }cos2x.d(e^x)$
$\Leftrightarrow e^xcos2x-\int\limits_{0}^{\pi }e^xd(cos2x)$
$\Leftrightarrow e^xcos2x+2\int\limits_{0}^{\pi }e^xsin2x.dx$
$\Leftrightarrow e^xcos2x+2e^xsin2x-2\int\limits_{0}^{\pi }e^xd(sin2x)$
$\Leftrightarrow e^xcos2x+2e^xsin2x+4\int\limits_{0}^{\pi }cos2x.e^xdx$
Ta có: $J= e^xcos2x+2e^xsin2x+4\int\limits_{0}^{\pi }cos2x.e^xdx$
$\Leftrightarrow\int\limits_{0}^{\pi }cos2x.e^xdx=e^xcos2x+2e^xsin2x+4\int\limits_{0}^{\pi }cos2x.e^xdx$
$\Leftrightarrow\int\limits_{0}^{\pi }cos2x.e^xdx=-\frac{1}{3}(e^xcos2x+2e^xsin2x)$
$\Rightarrow \int\limits_{0}^{\pi }cos^2x.e^xdx=\frac{1}{2}e^x|^{\pi }_0-\frac{1}{6}(e^xcos2x+2e^xsin2x)|^{\pi }_0$
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bình luận
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Giải giúp mình với!!
đây là đề cđ khối a 2013. mình cũng làm giống bạn mà không biết đáp án của bộ thì khác với đáp án trên, không biết nó có giống nhau không. Bạn xem rồi giải thích giúp mình với. Thắc mắc của mình là đây
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được thưởng
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Đăng nhập hàng ngày 21/01/2014
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sửa đổi
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tích phân
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Đặt $x=tanu$$\Rightarrow \begin{cases}dx=\frac{du}{cos^2u}\\ x^2\sqrt{x^2+1}=tan^2u.\sqrt{tan^2u+1}=\frac{tan^2u}{cosu} \end{cases}$Đổi cận nhé$\int\limits_{\frac{\pi }{4}}^{arctan2}\frac{cosu.du}{tan^2u.cos^2u}=\int\limits_{\frac{\pi }{4}}^{arctan2}\frac{du}{tan^2u.cosu}$$\int\limits_{\frac{\pi }{4}}^{arctan2}\frac{cosu.du}{sin^2u}=\int\limits_{\frac{\pi }{4}}^{arctan2}\frac{d(sinu)}{sin^2u}$$\Rightarrow -\int\limits_{\frac{\pi }{4}}^{arctan2}d(\frac{1}{sinu}) $
Đặt $x=tanu$$\Rightarrow \begin{cases}dx=\frac{du}{cos^2u}\\ x^2\sqrt{x^2+1}=tan^2u.\sqrt{tan^2u+1}=\frac{tan^2u}{cosu} \end{cases}$Đổi cận nhé$\int\limits_{\frac{\pi }{4}}^{arctan2}\frac{cosu.du}{tan^2u.cos^2u}=\int\limits_{\frac{\pi }{4}}^{arctan2}\frac{du}{tan^2u.cosu}$$\int\limits_{\frac{\pi }{4}}^{arctan2}\frac{cosu.du}{sin^2u}=\int\limits_{\frac{\pi }{4}}^{arctan2}\frac{d(sinu)}{sin^2u}$$\Rightarrow -\int\limits_{\frac{\pi }{4}}^{arctan2}d(\frac{1}{sinu}) =\int\limits_{arctan2}^{\frac{\pi }{4}}d(\frac{1}{sinu})$
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