Matematika/Integravimas keičiant kintamąjį

Integravimas keičiant kintamąjį:

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

$I=\int f(x)dx=\int f[\varphi (t)]{\varphi }^{\prime }(t)dt.$

Suintegrave, grįžtame prie senojo kintamojo.

2. Įvedę keitinį u=g(x), gauname:

$\int g(x)g^{\prime }(x)dx=\int udu.$

Pavyzdžiai

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

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

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:

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

Dabar jau aišku, kad reikia imti keitinį $t=\sin x,$ $dt=\cos xdx.$ 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 $d(\frac{x}{2})=\frac{1}{2}dx;dx=2d(\frac{x}{2}).$

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

• $\int \sqrt{1-x^2}𝖽x,$

Keitinys: $x=\sin t,𝖽x=\cos t𝖽t,t=\arcsin x$,

$\int \sqrt{1-{\sin }^{2}t}\cos t𝖽t=\int {\cos }^{2}t𝖽t=\int \frac{1+\cos (2t)}{2}𝖽t=$

$=\frac{1}{2}(\int 𝖽t+\frac{1}{2}\int \cos (2t)𝖽t)=\frac{1}{2}(\int 𝖽t+\frac{1}{2}\int \cos (2t)\frac{𝖽(2t)}{2})=\frac{t}{2}+\frac{\sin (2t)}{4}+C.$

Įstatę pakeistą kintamąjį gauname atsakymą:

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

• Apskaičiuosime $\int \cos (2x)dx.$ Šiuo atveju reikia pasirinkti labai paprastą keitinį $d(2x)=2dx,$ 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 $\int e^{\cos x}\sin xdx.$ Lengva numatyti, kad tas integralas apskaičiuojamas, naudojant keitinį $d(\cos x)=-\sin xdx.$ Tuomet $\sin xdx=-d(\cos x)$ ir

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

• 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 $dx=\frac{d(7x-9)}{7}.$ 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 $((2x{)}^4{)}^{\prime }=64x^3$. 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 $d(2x+1)=2dx;$ $dx=\frac{d(2x+1)}{2}.$

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

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

• $\underset{0}{\overset{\pi /2}{\int }}\frac{\cos x}{1+{\sin }^{2}x}dx=\underset{0}{\overset{\pi /2}{\int }}\frac{d(\sin x)}{1+{\sin }^{2}x}=\arctan (\sin x){|}_0^{\pi /2}=\arctan (\sin \frac{\pi }{2})-\arctan (\sin 0)=$

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

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

• $\underset{1}{\overset{e}{\int }}\frac{{\ln }^{2}x}{x}dx=\underset{1}{\overset{e}{\int }}{\ln }^{2}xd(\ln x)=\frac{{\ln }^{3}x}{3}{|}_1^e=\frac{{\ln }^{3}e}{3}-\frac{{\ln }^{3}1}{3}=\frac{1}{3}-0=\frac{1}{3},$

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

• $\underset{0}{\overset{1}{\int }}\frac{dx}{\sqrt{1-x}}=-\underset{0}{\overset{1}{\int }}\frac{d(1-x)}{\sqrt{1-x}}=-2\sqrt{1-x}{|}_0^1=-2\sqrt{1-1}+2\sqrt{1-0}=0+2=2,$

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

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

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

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

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

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

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

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

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

kur $t=\frac{1}{2}\cdot \arccos \frac{x}{a},2t=\arccos \frac{x}{a},\frac{x}{a}=\cos (2t),$ $x=a\cos (2t),$ $dx=-2a\sin (2t)dt,$ $dt=\frac{dx}{-2a\sin (2t)},$ d(2t)=2dt.

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

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

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

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

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

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

• $\int \frac{dx}{\sqrt{x^2-a^2}}.$ 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 $\sqrt{e^x+1}=t,e^x+1=t^2,e^x=t^2-1;e^{x}dx=2tdt,dx=\frac{2tdt}{e^x}=\frac{2tdt}{t^2-1}.$ Tada

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

Nuorodos

• http://integral-table.com/