Matematika/Integravimas keičiant kintamąjį

Integravimas keičiant kintamąjį:

1. Įvedę keitinį $x=\phi (t)$ , kur $\phi (t)$ - tolydžiai diferencijuojama funkcija, gauname:

I=\int f(x)dx=\int f[\phi (t)]\phi '(t)dt.

Suintegrave, grįžtame prie senojo kintamojo.

2. Įvedę keitinį u=g(x), gauname:

\int g(x)g'(x)dx=\int u\,du.

Pavyzdžiai

$\int {\frac {dx}{\sqrt {a^{2}-x^{2}}}}=\int {\frac {dx}{a{\sqrt {1-x^{2}/a^{2}}}}}=\int {\frac {{\mathsf {d}}(x/a)}{\sqrt {1-(x/a)^{2}}}}=\int {\frac {du}{\sqrt {1-u^{2}}}}=$

=\int {\frac {\cos(t)}{\sqrt {1-\sin ^{2}t}}}dt=\int {\frac {\cos t}{\sqrt {\cos ^{2}t}}}dt=\int {\frac {\cos t}{\cos t}}dt=\int dt=t+C=\arcsin u+C=\arcsin {\frac {x}{a}}+C,

kur d(x/a)=(dx)/a, dx=a*d(x/a), u=x/a,

u=\sin(t)

;

t=\arcsin u=\arcsin {\frac {x}{a}};

du=\cos(t)dt

.

$\int {\frac {dx}{a^{2}+x^{2}}}=\int {\frac {dx}{a^{2}(1+x^{2}/a^{2})}}={\frac {1}{a}}\int {\frac {d(x/a)}{1+(x/a)^{2}}}={\frac {1}{a}}\arctan {\frac {x}{a}}+C.$

$\int {\frac {dx}{x^{2}-a^{2}}}=\int {\frac {1}{2a}}({\frac {1}{x-a}}-{\frac {1}{x+a}})dx={\frac {1}{2a}}\int {\frac {dx}{x-a}}-{\frac {1}{2a}}\int {\frac {dx}{x+a}}=$

={\frac {1}{2a}}\int {\frac {d(x-a)}{x-a}}-{\frac {1}{2a}}\int {\frac {d(x+a)}{x+a}}={\frac {1}{2a}}\ln |x-a|-{\frac {1}{2a}}\ln |x+a|+C={\frac {1}{2a}}\ln |{\frac {x-a}{x+a}}|+C.

Apskaičiuosime $\int {\frac {dx}{\cos x}}.$ Kad būtų lengviau pasirinkti keitinį, integralą užrašysime šitaip:

\int {\frac {dx}{\cos x}}=\int {\frac {\cos x\;dx}{\cos ^{2}x}}=\int {\frac {\cos x\;dx}{1-\sin ^{2}x}}.

Dabar jau aišku, kad reikia imti keitinį

t=\sin x,

dt=\cos x\;dx.

Tada

\int {\frac {dx}{\cos x}}=\int {\frac {dt}{1-t^{2}}}={\frac {1}{2}}\ln |{\frac {1+t}{1-t}}|+C=\ln |{\text{tg}}\;({\frac {x}{2}}+{\frac {\pi }{4}})|+C.

Arba

\int {\frac {dx}{\cos x}}=\int {\frac {dt}{1-t^{2}}}=-\int {\frac {dt}{t^{2}-1}}=-{\frac {1}{2}}\ln |{\frac {t-1}{t+1}}|+C=-{\frac {1}{2}}\ln |{\frac {\sin x-1}{\sin x+1}}|+C=(?)=\ln |{\text{tg}}\;({\frac {x}{2}}+{\frac {\pi }{4}})|+C.

$\int x{\sqrt {1+x^{2}}}dx=\int x{\sqrt {1+x^{2}}}{\frac {d(1+x^{2})}{2x}}={\frac {1}{2}}\cdot {\frac {(1+x^{2})^{0.5+1}}{0.5+1}}+C={\frac {1}{3}}{\sqrt {(1+x^{2})^{3}}}+C,$

kur $d(1+x^{2})=2xdx;$ $dx={\frac {d(1+x^{2})}{2x}}.$

$\int {\frac {8}{x^{2}+4}}dx=$ $\int {\frac {2dx}{({\frac {x}{2}})^{2}+1}}=\int {\frac {4d({\frac {x}{2}})}{({\frac {x}{2}})^{2}+1}}=4\arctan {\frac {x}{2}},$ kur $d({\frac {x}{2}})={\frac {1}{2}}dx;\;dx=2d({\frac {x}{2}}).$

$\int \tan ^{2}x\sec ^{2}xdx=\int {\frac {\tan ^{2}x}{\cos ^{2}x}}dx=\int \tan ^{2}(x)\;{\mathsf {d}}(\tan x)={\frac {1}{3}}\tan ^{3}x+C,$ kur $d(\tan x)=\sec ^{2}xdx$ .

$\int 6e^{-2x}dx=6\int e^{-2x}{\frac {d(-2x)}{-2}}=-3\int e^{-2x}d(-2x)=-3e^{-2x}+C,$ kur $d(-2x)=-2dx;$ $dx={\frac {d(-2x)}{-2}}.$

$\int {\sqrt {1-x^{2}}}\;{\mathsf {d}}x,$

Keitinys: $x=\sin t,{\mathsf {d}}x=\cos t\;{\mathsf {d}}t,t=\arcsin x\quad$ ,

\int {\sqrt {1-\sin ^{2}t}}\;\cos t\;{\mathsf {d}}t=\int \cos ^{2}t\;{\mathsf {d}}t=\int {\frac {1+\cos(2t)}{2}}\;{\mathsf {d}}t=

={\frac {1}{2}}\left(\int {\mathsf {d}}t+{\frac {1}{2}}\int \cos(2t)\;{\mathsf {d}}t\right)={\frac {1}{2}}\left(\int {\mathsf {d}}t+{\frac {1}{2}}\int \cos(2t)\;{\frac {{\mathsf {d}}(2t)}{2}}\right)={\frac {t}{2}}+{\frac {\sin(2t)}{4}}+C.

Įstatę pakeistą kintamąjį gauname atsakymą:

\int {\sqrt {1-x^{2}}}\;{\mathsf {d}}x={\frac {\arcsin x}{2}}+{\frac {\sin(2\arcsin x)}{4}}+C.

Apskaičiuosime $\int \cos(2x)\;dx.$ Šiuo atveju reikia pasirinkti labai paprastą keitinį $d(2x)=2\;dx,$ todėl $dx={\frac {d(2x)}{2}}$ . Pasinaudoję tuo keitiniu, gauname

\int \cos(2x)=\int \cos(2x)\;{\frac {d(2x)}{2}}={\frac {1}{2}}\sin(2x)+C.

Apskaičiuosime $\int {\frac {dx}{x+a}}.$ Kadangi $dx=d(x+a)=1,$ tai

\int {\frac {dx}{x+a}}=\int {\frac {d(x+a)}{x+a}}=\ln |x+a|+C.

Apskaičiuosime $\int e^{\cos x}\sin x\;dx.$ Lengva numatyti, kad tas integralas apskaičiuojamas, naudojant keitinį $d(\cos x)=-\sin x\;dx.$ Tuomet $\sin x\;dx=-d(\cos x)$ ir

\int e^{\cos x}\sin x\;dx=-\int e^{\cos x}\;d(\cos x)=-e^{\cos x}+C.

Apskaičiuosime $\int {\frac {(\arctan x)^{100}}{1+x^{2}}}\;dx.$ Kadangi $d(\arctan x)={\frac {1}{1+x^{2}}},$ o dx=1, tai reiškinį ${\frac {(\arctan x)^{100}}{1+x^{2}}}$ galima perrašyt šitaip $(\arctan x)^{100}\;d(\arctan x).$ Todėl

\int {\frac {(\arctan x)^{100}}{1+x^{2}}}\;dx=\int (\arctan x)^{100}\;d(\arctan x)={\frac {(\arctan x)^{101}}{101}}+C.

Apskaičiuosime $\int (7x-9)^{2999}\;dx.$ Kadangi $d(7x-9)=7dx,$ tai $dx={d(7x-9) \over 7}.$ Tada

$\int (7x-9)^{2999}\;dx=\int (7x-9)^{2999}{\frac {d(7x-9)}{7}}={\frac {(7x-9)^{3000}}{21000}}+C.$

Apskaičiuosime $\int {\frac {x^{3}\;dx}{(2x)^{8}+1}}.$ Čia patogus keitinys $t=(2x)^{4}$ , $dt=64x^{3}$ dx, nes $((2x)^{4})'=64x^{3}$ . Tada

\int {\frac {x^{3}\;dx}{(2x)^{8}+1}}={\frac {1}{64}}\int {\frac {dt}{t^{2}+1}}={\frac {\arctan t}{64}}+C={\frac {\arctan(2x)^{4}}{64}}+C.

$\int x{\sqrt {x-2}}dx=\int (2+t^{2})t\cdot 2tdt=\int 4t^{2}+2t^{4}dt={\frac {4}{3}}t^{3}+{\frac {2}{5}}t^{5}+C={\frac {4}{3}}{\sqrt {(x-2)^{3}}}+{\frac {2}{5}}{\sqrt {(x-2)^{5}}}+C,$

kur ${\sqrt {x-2}}=t,$ $x-2=t^{2},$ $dx=d(x-2)=d(t^{2})=2tdt,$ $x=2+t^{2}.$

$\int {\frac {\cos x}{\sqrt {1+4\sin x}}}dx=\int {\frac {t/2}{t}}dt={\frac {1}{2}}\int dt={\frac {t}{2}}+C={\frac {\sqrt {1+4\sin x}}{2}}+C,$

kur ${\sqrt {1+4\sin x}}=t;$ $1+4\sin x=t^{2};$ $4\cos xdx=2tdt;$ $\cos xdx={\frac {t}{2}}dt.$

$\int {\frac {x^{2}dx}{\sqrt {3+x}}}=\int {\frac {(t^{2}-3)^{2}\cdot 2tdt}{t}}=2\int t^{4}dt-12\int t^{2}dt+18\int dt={\frac {2t^{5}}{5}}-{\frac {12t^{3}}{3}}+18t+C=$

={\frac {2{\sqrt {(3+x)^{5}}}}{5}}-4{\sqrt {(3+x)^{3}}}+18{\sqrt {3+x}}+C={\frac {2{\sqrt {3+x}}}{5}}[(3+x)^{2}-10(3+x)+45]+C=

={\frac {2{\sqrt {3+x}}}{5}}(x^{2}-4x+24)+C,

kur ${\sqrt {3+x}}=t;$ $3+x=t^{2};$ dx=2tdt; $x=t^{2}-3.$

$\int (2x+1)^{20}dx=\int (2x+1)^{20}{\frac {d(2x+1)}{2}}={\frac {(2x+1)^{21}}{42}}+C,$

kur $d(2x+1)=2dx;$ $dx={d(2x+1) \over 2}.$

$\int {\frac {(2\ln x+3)^{3}}{x}}dx=\int (2\ln x+3)^{3}{\frac {d(2\ln x+3)}{2}}={\frac {1}{8}}(2\ln x+3)^{4}+C,$

kur $d(2\ln x+3)={\frac {2}{x}}dx;$ ${\frac {dx}{x}}={\frac {d(2\ln x+3)}{2}}.$

$\int {\frac {dx}{\sin x\cdot \cos x}}=\int {\frac {dx}{\tan x\cdot \cos ^{2}x}}=\int {\frac {d(\tan x)}{\tan x}}=\ln |\tan x|+C.$
$\int {\frac {dx}{1+e^{x}}}=\int {\frac {dt}{t(t-1)}}=\int {\frac {dt}{t-1}}-\int {\frac {dt}{t}}=\ln |t-1|-\ln |t|+C=x-\ln(1+e^{x})+C,$

kur $1+e^{x}=t;$ $e^{x}=t-1;$ $x=\ln(t-1);$ dx=dt/(t-1).

$\int {\frac {dx}{\sin x}}=\int {\frac {dx}{2\sin(x/2)\cdot \cos(x/2)}}=\int {\frac {dt}{\sin t\cdot \cos t}}=\int {\frac {dt}{\tan t\cdot \cos ^{2}t}}=\int {\frac {d(\tan t)}{\tan t}}=$

=\ln |\tan t|+C=\ln |\tan {\frac {x}{2}}|+C,

kur x/2=t; dx/2=dt; dx=2dt.

$\int {\sqrt {a^{2}-x^{2}}}dx=\int {\sqrt {a^{2}(1-\cos ^{2}t)}}\cdot (-a\sin t)dt=-a^{2}\int \sin ^{2}tdt=-{\frac {a^{2}}{2}}\int (1-\cos(2t))dt=$

=-{\frac {a^{2}}{2}}t+{\frac {a^{2}}{4}}\sin(2t)+C=-{\frac {a^{2}}{2}}t+{\frac {a^{2}}{4}}\cdot 2\sin(t)\cdot \cos(t)+C=-{\frac {a^{2}}{2}}t+{\frac {a^{2}}{2}}\cdot {\sqrt {1-\cos ^{2}t}}\cdot \cos(t)+C=

=-{\frac {a^{2}}{2}}\cdot \arccos {\frac {x}{a}}+{\frac {a^{2}}{2}}\cdot {\sqrt {1-\cos ^{2}(\arccos {\frac {x}{a}})}}\cdot \cos(\arccos {\frac {x}{a}})+C=-{\frac {a^{2}}{2}}\arccos {\frac {x}{a}}+{\frac {x}{2}}{\sqrt {a^{2}-x^{2}}}+C,

kur $\sin(2t)=2\sin t\cos t=2{\sqrt {1-\cos ^{2}t}}\cdot \cos t=2{\sqrt {1-{\frac {x^{2}}{a^{2}}}}}\cdot {\frac {x}{a}}={\frac {2x}{a}}{\sqrt {\frac {a^{2}-x^{2}}{a^{2}}}}={\frac {2x}{a^{2}}}{\sqrt {a^{2}-x^{2}}};$

x=a\cos t;

{\frac {x}{a}}=\cos t;

t=\arccos {\frac {x}{a}};

dx=-a\sin(t)dt.

$\int {\frac {xdx}{\sqrt {1-x^{2}}}}=\int {\frac {-{\frac {1}{2}}d(1-x^{2})}{\sqrt {1-x^{2}}}}=-{\frac {1}{2}}{\frac {(1-x^{2})^{-0.5+1}}{-0.5+1}}+C=-{\sqrt {1-x^{2}}}+C,$

kur $d(1-x^{2})=-2xdx;$ $xdx={\frac {d(1-x^{2})}{-2}}.$

$\int _{0}^{\pi /2}{\frac {\cos x}{1+\sin ^{2}x}}dx=\int _{0}^{\pi /2}{\frac {d(\sin x)}{1+\sin ^{2}x}}=\arctan(\sin x)\vert _{0}^{\pi /2}=\arctan(\sin {\frac {\pi }{2}})-\arctan(\sin 0)=$

=\arctan 1-\arctan 0={\frac {\pi }{4}}-0={\frac {\pi }{4}},

kur $d(\sin x)=\cos xdx$ arba $dx={\frac {d(\sin x)}{\cos x}}.$

$\int _{1}^{e}{\frac {\ln ^{2}x}{x}}dx=\int _{1}^{e}\ln ^{2}xd(\ln x)={\frac {\ln ^{3}x}{3}}\vert _{1}^{e}={\frac {\ln ^{3}e}{3}}-{\frac {\ln ^{3}1}{3}}={\frac {1}{3}}-0={\frac {1}{3}},$

kur $d(\ln x)={\frac {dx}{x}}$ arba $dx=xd(\ln x).$

$\int _{0}^{1}{\frac {dx}{\sqrt {1-x}}}=-\int _{0}^{1}{\frac {d(1-x)}{\sqrt {1-x}}}=-2{\sqrt {1-x}}\vert _{0}^{1}=-2{\sqrt {1-1}}+2{\sqrt {1-0}}=0+2=2,$

kur d(1-x)=-dx; dx=-d(1-x).

$\int {\frac {2xdx}{1+x^{2}}}=\int {\frac {d(1+x^{2})}{1+x^{2}}}=\ln(1+x^{2})+C,$

kur $d(1+x^{2})=2xdx;$ $dx={\frac {d(1+x^{2})}{2x}}.$

$\int \cot xdx=\int {\frac {\cos x}{\sin x}}dx=\int {\frac {d(\sin x)}{\sin x}}=\ln |\sin x|+C,$

kur $d(\sin x)=\cos xdx.$

$\int \sin ^{3}x\cdot \cos xdx=\int \sin ^{3}xd(\sin x)={\frac {1}{4}}\sin ^{4}x+C,$

kur $d(\sin x)=\cos xdx.$

$\int {\frac {dx}{x{\sqrt {4x+1}}}}=\int {\frac {4dx}{(4x+1-1){\sqrt {4x+1}}}}=\int {\frac {4dx}{(({\sqrt {4x+1}})^{2}-1){\sqrt {4x+1}}}}=$

=\int {\frac {4{\sqrt {4x+1}}d({\sqrt {4x+1}})}{2(({\sqrt {4x+1}})^{2}-1){\sqrt {4x+1}}}}=\int {\frac {2d({\sqrt {4x+1}})}{({\sqrt {4x+1}})^{2}-1}}=-2\int {\frac {d({\sqrt {4x+1}})}{1-({\sqrt {4x+1}})^{2}}}=

$=-2\tanh ^{-1}({\sqrt {4x+1}})+C=-{\frac {2}{2}}\ln({\frac {1+{\sqrt {4x+1}}}{1-{\sqrt {4x+1}}}})+C=\ln({\frac {1-{\sqrt {4x+1}}}{1+{\sqrt {4x+1}}}})+C,$ kur $d({\sqrt {4x+1}})={\frac {2}{\sqrt {4x+1}}}dx;$ $dx={\frac {1}{2}}{\sqrt {4x+1}}d({\sqrt {4x+1}}).$

$\int {\frac {dx}{(a^{2}-x^{2})^{3/2}}}={\frac {1}{a^{2}}}\int {\frac {dt}{\cos ^{2}t}}={\frac {\tan t}{a^{2}}}+C={\frac {\sin t}{a^{2}{\sqrt {1-\sin ^{2}t}}}}+C={\frac {x}{a^{2}{\sqrt {a^{2}-x^{2}}}}}+C,$

kur $t=\arcsin {\frac {x}{a}},$ $x=a\sin t,$ $dx=a\cos tdt.$

$\int {\frac {\sin x}{\cos ^{2}x}}dx=\int {\frac {-d(\cos x)}{\cos ^{2}x}}=-\int (\cos x)^{-2}d(\cos x)=-{\frac {(\cos x)^{-2+1}}{-2+1}}+C={\frac {1}{\cos x}}+C=\sec x+C,$

kur $d(\cos x)=-\sin xdx;$ $dx={\frac {d(\cos x)}{-\sin x}}.$

$\int {\frac {x^{2}dx}{(1+x)^{4}}}=\int {\frac {(z-1)^{2}dz}{z^{4}}}=\int {\frac {z^{2}-2z+1}{z^{4}}}dz=\int {\frac {dz}{z^{2}}}-2\int {\frac {dz}{z^{3}}}+\int {\frac {dz}{z^{4}}}=-{\frac {1}{z}}+{\frac {1}{z^{2}}}-{\frac {1}{3z^{3}}}+C,$

kur $1+x=z;$ $dx=d(1+x)=dz;$ $x=z-1.$

$\int {\frac {dx}{\sqrt {1-x}}}=\int -{\frac {2t}{t}}dt=-2\int dt=-2t+C=-2{\sqrt {1-x}}+C,$ kur $t={\sqrt {1-x}};$ $x=1-t^{2};$ $dx=-2tdt.$
$\int {\frac {-2x}{(1+x^{2})^{2}}}dx=\int {\frac {-2x}{(1+x^{2})^{2}}}{\frac {d(1+x^{2})}{2x}}=\int {\frac {-1}{(1+x^{2})^{2}}}d(1+x^{2})=-{\frac {(1+x^{2})^{-2+1}}{-2+1}}+C=$

={\frac {1}{1+x^{2}}}+C,

kur $d(1+x^{2})=2xdx;$ $dx={\frac {d(1+x^{2})}{2x}}.$

$\int {\frac {\sin ^{3}x}{\cos x}}dx=\int {\frac {\sin x(1-\cos ^{2}x)}{\cos x}}dx=\int {\frac {\sin x(1-\cos ^{2}x)}{\cos x}}{\frac {d(\cos x)}{-\sin x}}=-\int {\frac {1-\cos ^{2}x}{\cos x}}d(\cos x)=$

$=-\int ({\frac {1}{\cos x}}-\cos x)d(\cos x)={\frac {\cos ^{2}x}{2}}-\ln \cos x+C,$ kur $d(\cos x)=-\sin xdx;$ $dx={\frac {d(\cos x)}{-\sin x}}.$

$\int {\frac {1}{\sin x-1}}dx={\frac {\sin x+1}{(\sin x-1)(\sin x+1)}}=\int {\frac {\sin x+1}{\sin ^{2}x-1}}dx=\int {\frac {\sin x+1}{-\cos ^{2}x}}dx=$

$=\int -{\frac {\sin x}{\cos ^{2}x}}-{\frac {1}{\cos ^{2}x}}dx=-\int {\frac {\sin x}{\cos ^{2}x}}{\frac {d(\cos x)}{-\sin x}}-\int {\frac {1}{\cos ^{2}x}}dx=\int {\frac {d(\cos x)}{\cos ^{2}x}}-\tan x+C_{2}=$ $=-{\frac {1}{\cos x}}-\tan x+C,$ kur $d(\cos x)=-\sin xdx;$ $dx={\frac {d(\cos x)}{-\sin x}}.$

$\int {\frac {x^{3}}{(x-1)^{2}}}dx=\int {(t+1)^{3} \over t^{2}}=\int (t+3+{\frac {3}{t}}+{\frac {1}{t^{2}}})dt={\frac {t^{2}}{2}}+3t+3\ln |t|-{\frac {1}{t}}+C=$

$={\frac {1}{2}}(x-1)^{2}+3(x-1)+3\ln |x-1|-{\frac {1}{x-1}}+C,$ kur $x-1=t;$ $x=t+1;$ $dx=dt.$

$\int {dx \over {\sqrt {x^{2}+a}}}=\int {dt \over t}=\ln |t|+C=\ln |{\sqrt {x^{2}+a}}+x|+C,$

kur ${\sqrt {x^{2}+a}}+x=t;\;dt=({\frac {x}{\sqrt {x^{2}+a}}}+1)dx={\frac {x+{\sqrt {x^{2}+a}}}{\sqrt {x^{2}+a}}}dx;\;dx={\frac {\sqrt {x^{2}+a}}{{\sqrt {x^{2}+a}}+x}}dt.$

$\int {e^{x}dx \over {\sqrt {4-e^{2x}}}}=\int {dt \over {\sqrt {4-t^{2}}}}=\arcsin {t \over 2}+C=\arcsin {e^{x} \over 2}+C,$ kur $e^{x}=t;$ $e^{x}dx=dt.$
$\int x{\sqrt {6-x^{2}}}dx=\int {\sqrt {6-x^{2}}}{d(6-x^{2}) \over -2}=-{1 \over 2}\cdot {(6-x^{2})^{\frac {3}{2}} \over {3 \over 2}}+C={1 \over 3}(x^{2}-6){\sqrt {6-x^{2}}}+C,$

$d(6-x^{2})=-2xdx;$ $dx=-d(6-x^{2})/2x.$

$\int {\sqrt {x+a \over a-x}}dx=-2a\int {\sqrt {a+a\cos(2t) \over a-a\cos(2t)}}\sin(2t)dt=-2a\int {\sqrt {(1+\cos(2t))^{2} \over 1-\cos ^{2}(2t)}}\sin(2t)dt=$

$=-2a\int {\sqrt {(2\cos ^{2}t)^{2} \over \sin ^{2}(2t)}}\sin(2t)dt=-2a\int {2\cos ^{2}t \over \sin(2t)}\sin(2t)dt=-4a\int \cos ^{2}tdt=$ $=-4a\int ({1 \over 2}+{1 \over 2}\cos(2t))dt=-2at-2a\int \cos(2t){d(2t) \over 2}=-2at-a\sin(2t)+C=$ $=-2at-a{\sqrt {1-\cos ^{2}(2t)}}+C=-a\arccos {x \over a}-a{\sqrt {1-({x \over a})^{2}}}+C=$ $=-a\arccos {x \over a}-a{\sqrt {a^{2}-x^{2} \over a^{2}}}+C=-a\arccos {x \over a}-{\sqrt {a^{2}-x^{2}}}+C,$

kur

t={\frac {1}{2}}\cdot \arccos {x \over a},\;2t=\arccos {x \over a},\;{\frac {x}{a}}=\cos(2t),

x=a\cos(2t),

dx=-2a\sin(2t)dt,

dt={\frac {dx}{-2a\sin(2t)}},

d(2t)=2dt.

$\int {dx \over (x^{2}+a^{2})^{3 \over 2}}=\int {a \over (a^{2}\tan ^{2}t+a^{2})^{3 \over 2}}{dt \over \cos ^{2}t}={1 \over a^{2}}\int {1 \over ({1 \over \cos ^{2}t})^{3 \over 2}}{dt \over \cos ^{2}t}={1 \over a^{2}}\int \cos ^{3}t{dt \over \cos ^{2}t}=$

$={1 \over a^{2}}\int \cos t\;dt={\sin t \over a^{2}}+C={\tan t \over a^{2}{\sqrt {1+\tan ^{2}t}}}+C={a\tan t \over a^{2}{\sqrt {a^{2}+a^{2}\tan ^{2}t}}}+C={x \over a^{2}{\sqrt {a^{2}+x^{2}}}}+C,$ kur $t=\arctan {x \over a},\;{x \over a}=\tan t,$ $x=a\tan t,$ $dx=a{dt \over \cos ^{2}t}.$

$\int {\sqrt {a^{2}-x^{2}}}\;dx.$ Darome keitinį $x=a\sin t.$ Tada $t=\arcsin {\frac {x}{a}},\;\;dx=a\cos t\;dt.$

\int {\sqrt {a^{2}-x^{2}}}\;dx=\int {\sqrt {a^{2}-a^{2}\sin ^{2}t}}\cdot a\cos t\;dt=a^{2}\int {\sqrt {1-\sin ^{2}}}\cos t\;dt=a^{2}\int \cos ^{2}t\;dt=

=a^{2}\int {\frac {1}{2}}(1+\cos 2t)dt={\frac {a^{2}}{2}}t+{\frac {a^{2}}{4}}\sin 2t+C={\frac {a^{2}}{2}}t+{\frac {a^{2}}{4}}\cdot 2\sin t\cos t+C=

={\frac {a^{2}}{2}}t+{\frac {a^{2}}{2}}\sin t{\sqrt {1-\sin ^{2}t}}+C={\frac {a^{2}}{2}}t+{\frac {1}{2}}a\sin t{\sqrt {a^{2}-a^{2}\sin ^{2}t}}+C={\frac {a^{2}}{2}}\arcsin {\frac {x}{a}}+{\frac {1}{2}}x{\sqrt {a^{2}-x^{2}}}+C.

$\int {\frac {dx}{\sqrt {x^{2}-a^{2}}}}.$ Imame keitinį $t=x+{\sqrt {x^{2}-a^{2}}}.$

Tada

dt=(1+{\frac {2x}{2{\sqrt {x^{2}-a^{2}}}}})dx={\frac {{\sqrt {x^{2}-a^{2}}}+x}{\sqrt {x^{2}-a^{2}}}}dx,\;\;dt={\frac {t}{\sqrt {x^{2}-a^{2}}}}dx,\;\;{\frac {dx}{\sqrt {x^{2}-a^{2}}}}={\frac {dt}{t}};

\int {\frac {dx}{\sqrt {x^{2}-a^{2}}}}=\int {\frac {dt}{t}}=\ln |t|+C=\ln |x+{\sqrt {x^{2}-a^{2}}}|+C.

$\int {\frac {dx}{\sqrt {e^{x}+1}}}.$ Imame keitinį ${\sqrt {e^{x}+1}}=t,\;\;e^{x}+1=t^{2},\;\;e^{x}=t^{2}-1;\;\;e^{x}\;dx=2t\;dt,\;\;dx={\frac {2t\;dt}{e^{x}}}={\frac {2t\;dt}{t^{2}-1}}.$ Tada

\int {\frac {dx}{\sqrt {e^{x}+1}}}=\int {\frac {\frac {2t\;dt}{t^{2}-1}}{t}}=\int {\frac {2\;dt}{t^{2}-1}}=\ln |{\frac {t-1}{t+1}}|+C=\ln |{\frac {{\sqrt {e^{x}+1}}-1}{{\sqrt {e^{x}+1}}+1}}|+C.

Nuorodos

http://integral-table.com/