I=\int_{}^{}dx/(cosx.cos(x+pi/4))
j=\int_{}^{}(sinx^3/cosx^4)dx
H=\int_{}^{}(dx/(\sqrt{x^2+K})
G=\int_{}^{}\sqrt{x^2+K}
*Với K là một số bất kì
đó là mấy bài có thể ra kiểm tra 15 phút của mình đó. mấy bạn giúp giùm mình nghen
[TEX]\begin{array}{l}I = \int {\frac{{dx}}{{\cos x.\cos \left( {x + \frac{\pi }{4}} \right)}}} = \int {\frac{{dx}}{{\cos x.\left( {\cos x.\frac{{\sqrt 2 }}{2} - \sin x.\frac{{\sqrt 2 }}{2}} \right)}}} \\ = \sqrt 2 \int {\frac{{dx}}{{{{\cos }^2}x\left( {1 - \tan x} \right)}}} = \sqrt 2 \int {\frac{{d\tan x}}{{1 - \tan x}} = } - \sqrt 2 \int {\frac{{d\left( {1 - \tan x} \right)}}{{1 - \tan x}} = } - \sqrt 2 \ln \left| {1 - \tan x} \right| \\ \end{array}[/TEX]
[TEX]\begin{array}{l}J = \int {\frac{{{{\sin }^3}x}}{{{{\cos }^4}x}}} dx = - \int {\frac{{{{\sin }^2}x}}{{{{\cos }^4}x}}} d\cos x = - \int {\frac{{1 - {{\cos }^2}x}}{{{{\cos }^4}x}}} d\cos x \\ = \int {\frac{{d\cos x}}{{{{\cos }^2}x}}} - \int {\frac{{d\cos x}}{{{{\cos }^4}x}}} = \frac{{ - 1}}{{2\cos x}} + \frac{1}{{3{{\cos }^3}x}} \\ \end{array}[/TEX]
[TEX]H = \int {\frac{{dx}}{{\sqrt {{x^2} + k} }}} = \ln \left| {x + \sqrt {{x^2} + k} } \right| + C[/TEX]
CM
[TEX]{\left( {\ln \left| {x + \sqrt {{x^2} + k} } \right| + C} \right)}\nolimits^' = \frac{{1 + \frac{x}{{\sqrt {{x^2} + k} }}}}{{x + \sqrt {{x^2} + k} }} = \frac{{\sqrt {{x^2} + k} + x}}{{\sqrt {{x^2} + k} \left( {x + \sqrt {{x^2} + k} } \right)}} = \frac{1}{{\sqrt {{x^2} + k} }}[/TEX]
G:
*
[TEX]k \ge 0[/TEX]
[TEX]\begin{array}{l}x = \sqrt k \tan t,t \in [0;\frac{\pi }{2}) \\ dx = \sqrt k .\frac{1}{{{{\cos }^2}t}}dt = \sqrt k \left( {{{\tan }^2}t + 1} \right)dt \\ \int {\sqrt {{x^2} + k} {\rm{ }}dx = } \int {\sqrt {k\left( {{{\tan }^2}t + 1} \right)} {\rm{ }}\sqrt k .\frac{1}{{{{\cos }^2}t}}dt} \\ \end{array}[/TEX]
[TEX]\begin{array}{l}= k\int {\frac{1}{{{{\cos }^3}t}}{\rm{ }}dt} = = k\int {\frac{{\cos t}}{{{{\cos }^4}t}}{\rm{ }}dt} = k\int {\frac{{d\sin t}}{{{{\left( {1 - {{\sin }^2}t} \right)}^2}}}{\rm{ }}} \\ = k\int {\frac{{du}}{{{{\left( {1 - {u^2}} \right)}^2}}}{\rm{ }}} = k\int {\frac{{du}}{{{{\left[ {\left( {1 - u} \right)\left( {1 + u} \right)} \right]}^2}}}{\rm{ }}} \left( {u = \sin t} \right) \\ = \frac{k}{4}\int {{{\left( {\frac{1}{{1 - u}} + \frac{1}{{1 + u}}} \right)}^2}du} \\ \end{array}[/TEX]
[TEX]\begin{array}{l}= \frac{k}{4}\int {\left[ {{{\left( {\frac{1}{{1 - u}}} \right)}^2} + {{\left( {\frac{1}{{1 + u}}} \right)}^2} + \frac{2}{{\left( {1 - u} \right)\left( {1 + u} \right)}}} \right]du} \\ = \frac{k}{4}\int {{{\left( {\frac{1}{{1 - u}}} \right)}^2}du + } \frac{k}{4}\int {{{\left( {\frac{1}{{1 + u}}} \right)}^2}du + } \frac{k}{2}\int {\frac{1}{{\left( {1 - u} \right)\left( {1 + u} \right)}}du} \\ = - \frac{k}{4}\int {{{\left( {1 - u} \right)}^{ - 2}}d\left( {1 - u} \right) + } \frac{k}{4}\int {{{\left( {1 + u} \right)}^{ - 2}}d(1 + u) + } \frac{k}{4}\int {\left( {\frac{1}{{1 - u}} + \frac{1}{{1 + u}}} \right)du} \\ = - \frac{k}{4}\int {{{\left( {1 - u} \right)}^{ - 2}}d\left( {1 - u} \right) + } \frac{k}{4}\int {{{\left( {1 + u} \right)}^{ - 2}}d(1 + u) + } \frac{k}{4}\int {\left( {\frac{1}{{1 - u}} + \frac{1}{{1 + u}}} \right)du} \\ = - \frac{k}{4}\int {{{\left( {1 - u} \right)}^{ - 2}}d\left( {1 - u} \right) + } \frac{k}{4}\int {{{\left( {1 + u} \right)}^{ - 2}}d(1 + u)} - \frac{k}{4}\int {\frac{1}{{1 - u}}d\left( {1 - u} \right)} + \frac{k}{4}\int {\frac{1}{{1 + u}}d\left( {1 + u} \right)} \\ = \frac{k}{4}.\frac{1}{{1 - u}} - \frac{k}{4}.\frac{1}{{1 + u}} - \frac{k}{4}\ln \left| {1 - u} \right| + \frac{k}{4}.\ln \left| {1 + u} \right| + C \\ = = \frac{k}{4}.\frac{1}{{1 - \sin t}} - \frac{k}{4}.\frac{1}{{1 + \sin t}} - \frac{k}{4}\ln \left| {1 - \sin t} \right| + \frac{k}{4}.\ln \left| {1 + \sin t} \right| + C \\ \end{array}[/TEX]
*
k<0 dễ
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