Integravimas keičiant kintamąjį:
1. Įvedę keitinį
, kur
- tolydžiai diferencijuojama funkcija, gauname:
![{\displaystyle I=\int f(x)dx=\int f[\phi (t)]\phi '(t)dt.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/857e4a1697f33de0922d525772bd426c3952e1a9)
Suintegrave, grįžtame prie senojo kintamojo.
2. Įvedę keitinį u=g(x), gauname:
![{\displaystyle \int g(x)g'(x)dx=\int u\,du.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/506ff13fd468ffb4a430983524f6b02790fa90b2)
Pavyzdžiai
![{\displaystyle \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}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/551df61c85cb2925a978ce41e58d2ebac1ceb710)
![{\displaystyle =\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,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c317c798060259b8072fff2b3f170b891190960e)
- kur d(x/a)=(dx)/a, dx=a*d(x/a), u=x/a,
;
.
![{\displaystyle \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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2296d339c12627e6c1dfe6a0ba7a199f43f801f1)
![{\displaystyle \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}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/727e31a51ee590d806ef4b38078d5fc132ec9a32)
![{\displaystyle ={\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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4a08f4ed8be83a9207b9335b2aeeb035586ed778)
- Apskaičiuosime
Kad būtų lengviau pasirinkti keitinį, integralą užrašysime šitaip:
![{\displaystyle \int {\frac {dx}{\cos x}}=\int {\frac {\cos x\;dx}{\cos ^{2}x}}=\int {\frac {\cos x\;dx}{1-\sin ^{2}x}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ca677298e1d3d88973b58fef08a207e8a8cc2186)
- Dabar jau aišku, kad reikia imti keitinį
Tada
Arba
![{\displaystyle \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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77ea5a332b597080b7d85cd9e4f6c5f586c04ee6)
![{\displaystyle \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,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/544b2c71e66c5230d39313c1d0666dff1afaf4d4)
kur
![{\displaystyle \int {\frac {8}{x^{2}+4}}dx=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fa75a6521ba9e42a8d4d2c109e68baddbb1a46b5)
kur ![{\displaystyle d({\frac {x}{2}})={\frac {1}{2}}dx;\;dx=2d({\frac {x}{2}}).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb06feed38fa6096c999d4dd38b33471b7439340)
kur
.
kur
![{\displaystyle dx={\frac {d(-2x)}{-2}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c35c3b2df08e53371064f2c6c15fa6898a2f0f8)
![{\displaystyle \int {\sqrt {1-x^{2}}}\;{\mathsf {d}}x,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b27f06e7fa785b3a416b2f5f5315bfe9c7210ad1)
Keitinys:
,
![{\displaystyle \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=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd5d682d4f7625e6b4a4e6931675b841041ecd2f)
![{\displaystyle ={\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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/966b1357a1b44b764f1ac83e3aa209f161d2075d)
Įstatę pakeistą kintamąjį gauname atsakymą:
![{\displaystyle \int {\sqrt {1-x^{2}}}\;{\mathsf {d}}x={\frac {\arcsin x}{2}}+{\frac {\sin(2\arcsin x)}{4}}+C.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/708f46461e0ec10926d2d38f50e6694571e051f1)
- Apskaičiuosime
Šiuo atveju reikia pasirinkti labai paprastą keitinį
todėl
. Pasinaudoję tuo keitiniu, gauname
![{\displaystyle \int \cos(2x)=\int \cos(2x)\;{\frac {d(2x)}{2}}={\frac {1}{2}}\sin(2x)+C.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a1a4340cec28713c2fe541b1fb2c828bc17ae9c6)
- Apskaičiuosime
Kadangi
tai
![{\displaystyle \int {\frac {dx}{x+a}}=\int {\frac {d(x+a)}{x+a}}=\ln |x+a|+C.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65818008c4a18ed2d6a9f1d4f8389d226fd71ad3)
- Apskaičiuosime
Lengva numatyti, kad tas integralas apskaičiuojamas, naudojant keitinį
Tuomet
ir
![{\displaystyle \int e^{\cos x}\sin x\;dx=-\int e^{\cos x}\;d(\cos x)=-e^{\cos x}+C.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18dbb8e5f3d58dce0d217477a7fb969b5c723623)
- Apskaičiuosime
Kadangi
o dx=1, tai reiškinį
galima perrašyt šitaip
Todėl
![{\displaystyle \int {\frac {(\arctan x)^{100}}{1+x^{2}}}\;dx=\int (\arctan x)^{100}\;d(\arctan x)={\frac {(\arctan x)^{101}}{101}}+C.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/847281cb4f88a7551a62c20db14bc7abc065cc30)
- Apskaičiuosime
Kadangi
tai
Tada
- Apskaičiuosime
Čia patogus keitinys
,
dx, nes
. Tada
![{\displaystyle \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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/44ac7fa046084a155057512f03cf4928f50b612c)
![{\displaystyle \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,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/867b9f27cd17d30eb74af4cbb31273980842a7ae)
kur
![{\displaystyle \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,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/036ecdcffb1478e1361ab6e347a3fc7f836f4025)
kur
![{\displaystyle \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=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e7c3f333c3ec3ffb8805a724196c8cdbe84120d)
![{\displaystyle ={\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=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1ea5b524d7f87ea4b75a3fa034a833bd9733d199)
![{\displaystyle ={\frac {2{\sqrt {3+x}}}{5}}(x^{2}-4x+24)+C,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/90d4704d05b2cb88e0b57e2c3c316589441d5693)
kur
dx=2tdt;
![{\displaystyle \int (2x+1)^{20}dx=\int (2x+1)^{20}{\frac {d(2x+1)}{2}}={\frac {(2x+1)^{21}}{42}}+C,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/016b8426befe1396551942d13e720433e5aa5613)
kur
![{\displaystyle \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,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2646e4ddc1cc457ea87534040f28d5c2813c5638)
kur
![{\displaystyle \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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/71a2737d0900ced4409de6c9f88f4fd10f36f2cf)
![{\displaystyle \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,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6759c58e1d9e9b2fcbabbd2f86a36c28535c9021)
kur
dx=dt/(t-1).
![{\displaystyle \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}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b54ec82e3406cdc712a2c1f114b872c95a86fbb)
![{\displaystyle =\ln |\tan t|+C=\ln |\tan {\frac {x}{2}}|+C,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/205c5aae070c6b15f877dccd4be5a47384705321)
kur x/2=t; dx/2=dt; dx=2dt.
![{\displaystyle \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=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4cb018224fb78d8d71c93cbcaa351d6e660048af)
![{\displaystyle =-{\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=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a2b137da95d8e37f2dea64dee977dea3de6dcc2)
![{\displaystyle =-{\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,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b50a7abc2cf5648fbab53e39db036d652a2466fb)
kur
![{\displaystyle dx=-a\sin(t)dt.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1b5b33444cdf49d158e24aac9e25b1a1458db6d4)
![{\displaystyle \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,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/934c7a85b6beec5f6e198f30c5566f3328ff285f)
kur
![{\displaystyle \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)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dad859bf81ccde843ef43e4a6c889c6909580ff5)
![{\displaystyle =\arctan 1-\arctan 0={\frac {\pi }{4}}-0={\frac {\pi }{4}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/38349f6b899829df59fa5272c7c42963087d2096)
kur
arba
![{\displaystyle \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}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf45abc28376ed86cbe51081e075a3adb640bb9e)
kur
arba
![{\displaystyle \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,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dc3f62f856dd688a82bca0a172c715cc8f857fe8)
kur d(1-x)=-dx; dx=-d(1-x).
![{\displaystyle \int {\frac {2xdx}{1+x^{2}}}=\int {\frac {d(1+x^{2})}{1+x^{2}}}=\ln(1+x^{2})+C,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c533d0304731be8f8588c850896975daa2c77aa)
kur
![{\displaystyle \int \cot xdx=\int {\frac {\cos x}{\sin x}}dx=\int {\frac {d(\sin x)}{\sin x}}=\ln |\sin x|+C,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/088679bce1edf030ab589a45f729ca36263a513f)
kur
![{\displaystyle \int \sin ^{3}x\cdot \cos xdx=\int \sin ^{3}xd(\sin x)={\frac {1}{4}}\sin ^{4}x+C,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f03a97dca6a66d854982c4243d0aa230efee1481)
kur
![{\displaystyle \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}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/143cbc38453b04c55fed61243f84fcf85ffd21f6)
![{\displaystyle =\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}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5700138b5bf73bcb5099524c254ce003b13960a5)
kur
![{\displaystyle \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,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b63625e876559b2c978b7212f09c53fb9857bb76)
kur
![{\displaystyle \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,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/47056b5187112aefd37ae430f651336cf6cb95c9)
kur
![{\displaystyle \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,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/45b2db488cc922017ce2a124b84f48a786a89365)
kur
kur
![{\displaystyle dx=-2tdt.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f5f99964e7cd522fdb38c9ea6fb8fb0ac6de231)
![{\displaystyle \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=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0c12299001dedb1fc80749e6f7184a02b4fcec11)
![{\displaystyle ={\frac {1}{1+x^{2}}}+C,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a0c6f657a0cf5636faf4eb18e20af080150bd218)
kur
![{\displaystyle \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)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d03e120a183094adc9ae155bc350e446b97b2261)
kur
![{\displaystyle \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=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e5c9e1d393ef7f6e1e01b90875489b5ea97957d)
kur
![{\displaystyle \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=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/739c53ea6f3d3a590824bc1f7b65915b2839e557)
kur
![{\displaystyle \int {dx \over {\sqrt {x^{2}+a}}}=\int {dt \over t}=\ln |t|+C=\ln |{\sqrt {x^{2}+a}}+x|+C,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/570c0a390d0efea7c2064a0ad771aa941b06c955)
kur
kur
![{\displaystyle e^{x}dx=dt.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/364bea2f76d9ba887c8473381e4c1a4135131160)
![{\displaystyle \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,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/29eb3dd1081d43811189cb8ee2fe015237d41c61)
![{\displaystyle \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=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6c48c167667877940caabaea2176fe78a08f450)
- kur
d(2t)=2dt.
![{\displaystyle \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}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b206d2e0bf30222d92f1286fe1716111abb95618)
kur
Darome keitinį
Tada ![{\displaystyle t=\arcsin {\frac {x}{a}},\;\;dx=a\cos t\;dt.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6fb8e4ae7aa46ab42645715fab79d32ffdc0de24)
![{\displaystyle \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=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e2e3d5dbb838a4b9c49975ee0a0e8645ddd0a1c1)
![{\displaystyle =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=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/174e8a5275b346109e0f73c7cc7ef29c1c2f0ff8)
![{\displaystyle ={\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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/146f68cbc7b750ff7e98d0f152f56d2a59c6b1a6)
Imame keitinį ![{\displaystyle t=x+{\sqrt {x^{2}-a^{2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98611c1e8ee96a33836ceb494f7c1b69d2acb60f)
- Tada
![{\displaystyle 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}};}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dbe9bbe5f58cdb1b53d2e0bf6fc4103e1625f096)
![{\displaystyle \int {\frac {dx}{\sqrt {x^{2}-a^{2}}}}=\int {\frac {dt}{t}}=\ln |t|+C=\ln |x+{\sqrt {x^{2}-a^{2}}}|+C.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be39f9e5ee2538666ef8f29e31d36e3bac93002b)
Imame keitinį
Tada
![{\displaystyle \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.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c01948b24156caf17494238784dba69916b6bd5)