I. Integralai
∫
sin
m
x
cos
n
x
d
x
,
{\displaystyle \int \sin ^{m}x\cos ^{n}xdx,}
m , n - sveikieji skaičiai, suvedami į integralą su binominiu diferencialu ir integruojami tik 3 atvejais:
1)n nelyginis;
2)m nelyginis;
3)m+n lyginis. Jei n nelyginis , taikome keitinį
sin
x
=
t
,
{\displaystyle \sin x=t,}
m nelyginis , taikome keitinį
cos
x
=
t
;
{\displaystyle \cos x=t;}
m
+
n
{\displaystyle m+n}
lyginis , keičiame
tan
x
=
t
;
{\displaystyle \tan x=t;}
sin
x
=
t
1
+
t
2
;
cos
x
=
1
1
+
t
2
;
d
x
=
d
t
1
+
t
2
.
{\displaystyle \sin x={\frac {t}{\sqrt {1+t^{2}}}};\;\cos x={\frac {1}{\sqrt {1+t^{2}}}};\;dx={\frac {dt}{1+t^{2}}}.}
Pavyzdžiai
∫
d
x
5
cos
2
x
+
9
sin
2
x
.
tan
x
=
t
;
sin
x
=
t
1
+
t
2
;
cos
x
=
1
1
+
t
2
;
d
x
=
d
t
1
+
t
2
.
{\displaystyle \int {\frac {dx}{5\cos ^{2}x+9\sin ^{2}x}}.\;\tan x=t;\;\sin x={\frac {t}{\sqrt {1+t^{2}}}};\;\cos x={\frac {1}{\sqrt {1+t^{2}}}};\;dx={\frac {dt}{1+t^{2}}}.}
∫
d
t
1
+
t
2
5
(
1
1
+
t
2
)
2
+
9
(
t
1
+
t
2
)
2
=
∫
d
t
5
+
9
t
2
=
1
3
5
arctan
3
t
5
+
C
.
{\displaystyle \int {\frac {\frac {dt}{1+t^{2}}}{5({\frac {1}{\sqrt {1+t^{2}}}})^{2}+9({\frac {t}{\sqrt {1+t^{2}}}})^{2}}}=\int {\frac {dt}{5+9t^{2}}}={\frac {1}{3{\sqrt {5}}}}\arctan {\frac {3t}{\sqrt {5}}}+C.}
∫
cos
4
x
sin
2
x
d
x
=
∫
(
1
−
sin
2
x
)
2
sin
2
x
d
x
=
∫
(
1
sin
2
x
−
2
+
sin
2
x
)
d
x
=
−
cot
x
−
2
x
+
1
2
∫
(
1
−
cos
(
2
x
)
)
d
x
=
{\displaystyle \int {\frac {\cos ^{4}x}{\sin ^{2}x}}dx=\int {\frac {(1-\sin ^{2}x)^{2}}{\sin ^{2}x}}dx=\int ({\frac {1}{\sin ^{2}x}}-2+\sin ^{2}x)dx=-\cot x-2x+{\frac {1}{2}}\int (1-\cos(2x))dx=}
=
−
cot
x
−
3
x
2
−
sin
(
2
x
)
4
+
C
.
{\displaystyle =-\cot x-{\frac {3x}{2}}-{\frac {\sin(2x)}{4}}+C.}
∫
sin
2
x
cos
6
x
d
x
.
{\displaystyle \int {\frac {\sin ^{2}x}{\cos ^{6}x}}dx.}
m ir n lyginiai,
m
=
2
,
{\displaystyle m=2,}
n
=
−
6
,
{\displaystyle n=-6,}
m
+
n
=
−
4
{\displaystyle m+n=-4}
tan
x
=
t
;
{\displaystyle \tan x=t;}
Nepavyko apdoroti (SVG (MathML gali būti įjungtas per naršyklės įskiepį): Netinkamas atsakas ("Math extension cannot connect to Restbase.") iš serverio "http://localhost:6011/lt.wikibooks.org/v1/":): {\displaystyle \sin^2 x=\frac{t^2}{1+t^2};}
cos
x
=
1
1
+
t
2
;
{\displaystyle \cos x={\frac {1}{\sqrt {1+t^{2}}}};}
cos
6
x
=
1
(
1
+
t
2
)
3
;
{\displaystyle \cos ^{6}x={\frac {1}{(1+t^{2})^{3}}};}
d
x
=
d
t
1
+
t
2
.
{\displaystyle dx={\frac {dt}{1+t^{2}}}.}
∫
sin
2
x
cos
6
x
d
x
=
∫
t
2
1
+
t
2
⋅
d
t
1
+
t
2
1
(
1
+
t
2
)
3
=
∫
t
2
d
t
(
1
+
t
2
)
2
1
(
1
+
t
2
)
3
=
∫
t
2
(
1
+
t
2
)
2
⋅
(
1
+
t
2
)
3
d
t
=
∫
t
2
(
1
+
t
2
)
d
t
=
t
3
3
+
t
5
5
+
C
=
tan
3
x
3
+
tan
5
x
5
+
C
.
{\displaystyle \int {\frac {\sin ^{2}x}{\cos ^{6}x}}dx=\int {\frac {{\frac {t^{2}}{1+t^{2}}}\cdot {\frac {dt}{1+t^{2}}}}{\frac {1}{(1+t^{2})^{3}}}}=\int {\frac {\frac {t^{2}\;dt}{(1+t^{2})^{2}}}{\frac {1}{(1+t^{2})^{3}}}}=\int {\frac {t^{2}}{(1+t^{2})^{2}}}\cdot (1+t^{2})^{3}dt=\int t^{2}(1+t^{2})dt={\frac {t^{3}}{3}}+{\frac {t^{5}}{5}}+C={\frac {\tan ^{3}x}{3}}+{\frac {\tan ^{5}x}{5}}+C.}
Apskaičiuosime integralą
∫
sin
x
cos
x
d
x
sin
4
x
+
cos
4
x
.
{\displaystyle \int {\frac {\sin x\cos x\;dx}{\sin ^{4}x+\cos ^{4}x}}.}
Kadangi pointegralinė funkcija nekeičia reikšmės, kai kartu keičiami
sin
x
{\displaystyle \sin x}
cos
x
{\displaystyle \cos x}
t
=
tan
x
,
{\displaystyle t=\tan x,}
∫
sin
x
cos
x
d
x
sin
4
x
+
cos
4
x
=
∫
t
1
+
t
2
1
1
+
t
2
d
t
1
+
t
2
t
4
(
1
+
t
2
)
2
+
1
(
1
+
t
2
)
2
=
∫
t
(
1
+
t
2
)
2
d
t
t
4
+
1
(
1
+
t
2
)
2
=
∫
t
d
t
t
4
+
1
=
1
2
∫
d
(
t
2
)
(
t
2
)
2
+
1
=
1
2
arctan
(
t
2
)
+
C
=
1
2
arctan
(
tan
2
x
)
+
C
,
{\displaystyle \int {\frac {\sin x\cos x\;dx}{\sin ^{4}x+\cos ^{4}x}}=\int {\frac {{\frac {t}{\sqrt {1+t^{2}}}}{\frac {1}{\sqrt {1+t^{2}}}}{\frac {dt}{1+t^{2}}}}{{\frac {t^{4}}{(1+t^{2})^{2}}}+{\frac {1}{(1+t^{2})^{2}}}}}=\int {\frac {{\frac {t}{(1+t^{2})^{2}}}dt}{\frac {t^{4}+1}{(1+t^{2})^{2}}}}=\int {\frac {t\;dt}{t^{4}+1}}={\frac {1}{2}}\int {\frac {d(t^{2})}{(t^{2})^{2}+1}}={\frac {1}{2}}\arctan(t^{2})+C={\frac {1}{2}}\arctan(\tan ^{2}x)+C,}
kur
sin
4
x
=
t
4
(
1
+
t
2
)
2
,
cos
4
x
=
1
(
1
+
t
2
)
2
,
d
x
=
d
t
1
+
t
2
.
{\displaystyle \sin ^{4}x={\frac {t^{4}}{(1+t^{2})^{2}}},\;\;\cos ^{4}x={\frac {1}{(1+t^{2})^{2}}},\;\;dx={\frac {dt}{1+t^{2}}}.}
II. Integralai
∫
sin
x
cos
x
d
x
{\displaystyle \int \sin x\cos xdx}
tan
x
2
=
t
.
{\displaystyle \tan {\frac {x}{2}}=t.}
sin
x
=
2
t
1
+
t
2
;
cos
x
=
1
−
t
2
1
+
t
2
;
d
x
=
2
d
t
1
+
t
2
.
{\displaystyle \sin x={\frac {2t}{1+t^{2}}};\;\cos x={\frac {1-t^{2}}{1+t^{2}}};\;dx={\frac {2\;dt}{1+t^{2}}}.}
Pavyzdžiai
∫
d
x
3
+
sin
x
+
cos
x
.
{\displaystyle \int {\frac {dx}{3+\sin x+\cos x}}.}
tan
x
2
=
t
;
{\displaystyle \tan {\frac {x}{2}}=t;}
x
2
=
arctan
t
;
{\displaystyle {\frac {x}{2}}=\arctan t;}
x
=
2
arctan
t
;
{\displaystyle x=2\arctan t;}
sin
x
=
2
t
1
+
t
2
;
{\displaystyle \sin x={\frac {2t}{1+t^{2}}};}
cos
x
=
1
−
t
2
1
+
t
2
;
{\displaystyle \cos x={\frac {1-t^{2}}{1+t^{2}}};}
d
x
=
d
(
2
arctan
t
)
=
2
d
t
1
+
t
2
.
{\displaystyle dx=d(2\arctan t)={\frac {2\;dt}{1+t^{2}}}.}
∫
d
x
3
+
sin
x
+
cos
x
=
∫
2
d
t
1
+
t
2
3
+
2
t
1
+
t
2
+
1
−
t
2
1
+
t
2
=
∫
2
d
t
1
+
t
2
2
(
t
2
+
t
+
2
)
1
+
t
2
=
∫
d
t
t
2
+
t
+
2
=
∫
d
t
(
t
+
1
2
)
2
+
7
4
=
{\displaystyle \int {\frac {dx}{3+\sin x+\cos x}}=\int {\frac {\frac {2\;dt}{1+t^{2}}}{3+{\frac {2t}{1+t^{2}}}+{\frac {1-t^{2}}{1+t^{2}}}}}=\int {\frac {\frac {2\;dt}{1+t^{2}}}{\frac {2(t^{2}+t+2)}{1+t^{2}}}}=\int {\frac {dt}{t^{2}+t+2}}=\int {\frac {dt}{(t+{\frac {1}{2}})^{2}+{\frac {7}{4}}}}=}
=
2
7
arctan
2
t
+
1
7
+
C
=
2
7
arctan
2
tan
x
2
+
1
7
+
C
.
{\displaystyle ={\frac {2}{\sqrt {7}}}\arctan {\frac {2t+1}{\sqrt {7}}}+C={\frac {2}{\sqrt {7}}}\arctan {\frac {2\tan {\frac {x}{2}}+1}{\sqrt {7}}}+C.}
∫
cot
x
d
x
1
−
sin
x
−
cos
x
=
∫
cos
x
d
x
sin
x
(
1
−
sin
x
−
cos
x
)
=
∫
1
−
t
2
1
+
t
2
⋅
2
d
t
1
+
t
2
2
t
1
+
t
2
(
1
−
2
t
1
+
t
2
−
1
−
t
2
1
+
t
2
)
=
{\displaystyle \int {\frac {\cot x\;dx}{1-\sin x-\cos x}}=\int {\frac {\cos x\;dx}{\sin x(1-\sin x-\cos x)}}=\int {\frac {{\frac {1-t^{2}}{1+t^{2}}}\cdot {\frac {2\;dt}{1+t^{2}}}}{{\frac {2t}{1+t^{2}}}(1-{\frac {2t}{1+t^{2}}}-{\frac {1-t^{2}}{1+t^{2}}})}}=}
=
∫
2
(
1
−
t
2
)
d
t
(
1
+
t
2
)
2
2
t
+
2
t
3
−
4
t
2
−
2
t
+
2
t
3
(
1
+
t
2
)
2
=
∫
2
(
1
−
t
)
(
1
+
t
)
d
t
4
t
3
−
4
t
2
=
∫
2
(
1
−
t
)
(
1
+
t
)
d
t
−
4
t
2
(
1
−
t
)
=
−
1
2
∫
(
1
t
2
+
1
t
)
d
t
=
1
2
(
cot
x
2
−
ln
|
tan
x
2
|
)
+
C
,
{\displaystyle =\int {\frac {\frac {2(1-t^{2})dt}{(1+t^{2})^{2}}}{\frac {2t+2t^{3}-4t^{2}-2t+2t^{3}}{(1+t^{2})^{2}}}}=\int {\frac {2(1-t)(1+t)dt}{4t^{3}-4t^{2}}}=\int {\frac {2(1-t)(1+t)dt}{-4t^{2}(1-t)}}=-{\frac {1}{2}}\int ({\frac {1}{t^{2}}}+{\frac {1}{t}})dt={\frac {1}{2}}(\cot {\frac {x}{2}}-\ln |\tan {\frac {x}{2}}|)+C,}
tan
x
2
=
t
;
sin
x
=
2
t
1
+
t
2
;
cos
x
=
1
−
t
2
1
+
t
2
;
d
x
=
2
d
t
1
+
t
2
.
{\displaystyle \tan {\frac {x}{2}}=t;\;\sin x={\frac {2t}{1+t^{2}}};\;\cos x={\frac {1-t^{2}}{1+t^{2}}};\;dx={\frac {2\;dt}{1+t^{2}}}.}
III. Integralams
∫
R
(
x
,
a
2
−
x
2
)
d
x
=
∫
x
±
1
(
a
2
−
x
2
)
±
1
d
x
{\displaystyle \int R(x,\;{\sqrt {a^{2}-x^{2}}})dx=\int x^{\pm 1}({\sqrt {a^{2}-x^{2}}})^{\pm 1}dx}
x
=
a
sin
t
,
{\displaystyle x=a\sin t,}
x
=
a
tan
t
{\displaystyle x=a\tan t}
x
=
a
sin
t
.
{\displaystyle x={\frac {a}{\sin t}}.}
Pavyzdžiai
∫
x
9
−
x
2
d
x
=
∫
3
sin
t
9
−
9
sin
2
t
⋅
3
cos
t
d
t
=
27
∫
sin
t
1
−
sin
2
t
⋅
cos
t
d
t
=
{\displaystyle \int x{\sqrt {9-x^{2}}}dx=\int 3\sin t{\sqrt {9-9\sin ^{2}t}}\cdot 3\cos t\;dt=27\int \sin t{\sqrt {1-\sin ^{2}t}}\cdot \cos t\;dt=}
=
27
∫
sin
t
cos
t
cos
t
d
t
=
−
27
∫
cos
2
t
d
(
cos
t
)
=
−
27
cos
3
t
3
+
C
=
−
9
(
1
−
sin
2
t
)
1
−
sin
2
t
+
C
=
{\displaystyle =27\int \sin t\cos t\cos t\;dt=-27\int \cos ^{2}t\;d(\cos t)=-27{\frac {\cos ^{3}t}{3}}+C=-9(1-\sin ^{2}t){\sqrt {1-\sin ^{2}t}}+C=}
=
−
9
(
1
−
x
2
3
2
)
1
−
x
2
3
2
+
C
=
−
1
3
(
9
−
x
2
)
9
−
x
2
+
C
,
{\displaystyle =-9(1-{\frac {x^{2}}{3^{2}}}){\sqrt {1-{\frac {x^{2}}{3^{2}}}}}+C=-{\frac {1}{3}}(9-x^{2}){\sqrt {9-x^{2}}}+C,}
3
sin
t
=
x
;
{\displaystyle 3\sin t=x;}
sin
t
=
x
3
;
{\displaystyle \sin t={\frac {x}{3}};}
d
x
=
3
cos
t
d
t
.
{\displaystyle dx=3\cos tdt.}
∫
d
x
x
a
2
+
x
2
=
∫
a
d
t
a
cos
2
t
⋅
tan
t
a
2
+
a
2
tan
2
t
=
∫
a
d
t
a
2
cos
2
t
⋅
tan
t
1
+
tan
2
t
=
∫
cos
t
d
t
a
cos
2
t
⋅
tan
t
=
1
a
∫
d
t
sin
t
=
{\displaystyle \int {\frac {dx}{x{\sqrt {a^{2}+x^{2}}}}}=\int {\frac {a\;dt}{a\cos ^{2}t\cdot \tan t{\sqrt {a^{2}+a^{2}\tan ^{2}t}}}}=\int {\frac {a\;dt}{a^{2}\cos ^{2}t\cdot \tan t{\sqrt {1+\tan ^{2}t}}}}=\int {\frac {\cos t\;dt}{a\cos ^{2}t\cdot \tan t}}={\frac {1}{a}}\int {\frac {dt}{\sin t}}=}
=
1
a
ln
|
1
−
cos
t
sin
t
|
+
C
=
1
a
ln
|
1
sin
t
−
cot
t
|
+
C
=
1
a
ln
|
a
2
+
x
2
−
a
x
|
+
C
,
{\displaystyle ={\frac {1}{a}}\ln |{\frac {1-\cos t}{\sin t}}|+C={\frac {1}{a}}\ln |{\frac {1}{\sin t}}-\cot t|+C={\frac {1}{a}}\ln |{\frac {{\sqrt {a^{2}+x^{2}}}-a}{x}}|+C,}
kur
x
=
a
tan
t
;
{\displaystyle x=a\tan t;}
d
x
=
a
cos
2
t
;
{\displaystyle dx={\frac {a}{\cos ^{2}t}};}
tan
t
=
x
a
;
{\displaystyle \tan t={\frac {x}{a}};}
cot
t
=
a
x
;
{\displaystyle \cot t={\frac {a}{x}};}
1
sin
t
=
1
+
cot
2
t
=
a
2
+
x
2
x
;
{\displaystyle {\frac {1}{\sin t}}={\sqrt {1+\cot ^{2}t}}={\frac {\sqrt {a^{2}+x^{2}}}{x}};}
1
+
tan
2
t
=
1
cos
2
t
.
{\displaystyle 1+\tan ^{2}t={\frac {1}{\cos ^{2}t}}.}
keisti
∫
R
(
sin
x
,
cos
x
)
d
x
=
∫
R
(
2
t
1
+
t
2
,
1
−
t
2
1
+
t
2
)
2
d
t
1
+
t
2
.
{\displaystyle \int R(\sin x,\cos x){\mathsf {d}}x=\int R\left({\frac {2t}{1+t^{2}}},\;{\frac {1-t^{2}}{1+t^{2}}}\right){\frac {2{\mathsf {d}}t}{1+t^{2}}}.}
Imamas keitinys
tan
x
2
=
t
,
−
π
<
x
<
π
.
{\displaystyle \tan {\frac {x}{2}}=t,\quad -\pi <x<\pi .}
Funkcijos
sin
(
x
)
{\displaystyle \sin(x)}
cos
(
x
)
{\displaystyle \cos(x)}
tan
x
2
{\displaystyle \tan {\frac {x}{2}}}
t .
sin
x
=
2
sin
x
2
cos
x
2
1
=
2
sin
x
2
cos
x
2
sin
2
x
2
+
cos
2
x
2
=
2
(
sin
x
2
cos
x
2
)
/
cos
2
x
2
(
sin
2
x
2
+
cos
2
x
2
)
/
cos
2
x
2
=
2
tan
x
2
tan
2
x
2
+
1
=
2
t
1
+
t
2
,
{\displaystyle \sin x={\frac {2\sin {\frac {x}{2}}\cos {\frac {x}{2}}}{1}}={\frac {2\sin {\frac {x}{2}}\cos {\frac {x}{2}}}{\sin ^{2}{\frac {x}{2}}+\cos ^{2}{\frac {x}{2}}}}={\frac {2(\sin {\frac {x}{2}}\cos {\frac {x}{2}})/\cos ^{2}{\frac {x}{2}}}{(\sin ^{2}{\frac {x}{2}}+\cos ^{2}{\frac {x}{2}})/\cos ^{2}{\frac {x}{2}}}}={\frac {2\tan {\frac {x}{2}}}{\tan ^{2}{\frac {x}{2}}+1}}={\frac {2t}{1+t^{2}}},}
cos
x
=
cos
2
x
2
−
sin
2
x
2
1
=
cos
2
x
2
−
sin
2
x
2
sin
2
x
2
+
cos
2
x
2
=
(
cos
2
x
2
−
sin
2
x
2
)
/
cos
2
x
2
(
sin
2
x
2
+
cos
2
x
2
)
/
cos
2
x
2
=
1
−
tan
2
x
2
tan
2
x
2
+
1
=
1
−
t
2
1
+
t
2
.
{\displaystyle \cos x={\frac {\cos ^{2}{\frac {x}{2}}-\sin ^{2}{\frac {x}{2}}}{1}}={\frac {\cos ^{2}{\frac {x}{2}}-\sin ^{2}{\frac {x}{2}}}{\sin ^{2}{\frac {x}{2}}+\cos ^{2}{\frac {x}{2}}}}={\frac {(\cos ^{2}{\frac {x}{2}}-\sin ^{2}{\frac {x}{2}})/\cos ^{2}{\frac {x}{2}}}{(\sin ^{2}{\frac {x}{2}}+\cos ^{2}{\frac {x}{2}})/\cos ^{2}{\frac {x}{2}}}}={\frac {1-\tan ^{2}{\frac {x}{2}}}{\tan ^{2}{\frac {x}{2}}+1}}={\frac {1-t^{2}}{1+t^{2}}}.}
Toliau
x
2
=
arctan
t
,
x
=
2
arctan
t
,
d
x
=
2
d
t
1
+
t
2
.
{\displaystyle {\frac {x}{2}}=\arctan t,\;\;x=2\arctan t,\quad {\mathsf {d}}x={\frac {2{\mathsf {d}}t}{1+t^{2}}}.}
∫
d
x
sin
x
=
∫
2
d
t
1
+
t
2
2
t
1
+
t
2
=
d
t
t
=
ln
|
t
|
+
C
=
ln
|
tan
x
2
|
+
C
.
{\displaystyle \int {\frac {dx}{\sin x}}=\int {\frac {\frac {2{\mathsf {d}}t}{1+t^{2}}}{\frac {2t}{1+t^{2}}}}={\frac {dt}{t}}=\ln |t|+C=\ln \left|\tan {\frac {x}{2}}\right|+C.}
∫
d
x
1
+
sin
x
=
∫
2
d
t
1
+
t
2
1
+
2
t
1
+
t
2
=
∫
2
d
t
(
1
+
t
2
)
1
+
t
2
+
2
t
1
+
t
2
=
∫
2
d
t
t
2
+
2
t
+
1
=
∫
2
d
t
(
t
+
1
)
2
=
{\displaystyle \int {\frac {{\mathsf {d}}x}{1+\sin x}}=\int {\frac {\frac {2{\mathsf {d}}t}{1+t^{2}}}{1+{\frac {2t}{1+t^{2}}}}}=\int {\frac {2{\mathsf {d}}t}{(1+t^{2}){\frac {1+t^{2}+2t}{1+t^{2}}}}}=\int {\frac {2{\mathsf {d}}t}{t^{2}+2t+1}}=\int {\frac {2{\mathsf {d}}t}{(t+1)^{2}}}=}
=
∫
2
d
(
t
+
1
)
(
t
+
1
)
2
=
−
2
t
+
1
+
C
=
−
2
tan
x
2
+
1
+
C
1
.
{\displaystyle =\int {\frac {2{\mathsf {d}}(t+1)}{(t+1)^{2}}}=-{\frac {2}{t+1}}+C=-{\frac {2}{\tan {\frac {x}{2}}+1}}+C_{1}.}
∫
d
x
9
+
8
cos
x
+
sin
x
=
∫
2
d
t
(
1
+
t
2
)
(
9
+
8
⋅
1
−
t
2
1
+
t
2
+
2
t
1
+
t
2
)
=
∫
2
d
t
(
1
+
t
2
)
⋅
9
(
1
+
t
2
)
+
8
(
1
−
t
2
)
+
2
t
1
+
t
2
=
{\displaystyle \int {\frac {{\mathsf {d}}x}{9+8\cos x+\sin x}}=\int {\frac {2{\mathsf {d}}t}{(1+t^{2})(9+8\cdot {\frac {1-t^{2}}{1+t^{2}}}+{\frac {2t}{1+t^{2}}})}}=\int {\frac {2{\mathsf {d}}t}{(1+t^{2})\cdot {\frac {9(1+t^{2})+8(1-t^{2})+2t}{1+t^{2}}}}}=}
=
∫
2
d
t
9
(
1
+
t
2
)
+
8
(
1
−
t
2
)
+
2
t
=
2
∫
d
t
9
+
9
t
2
+
8
−
8
t
2
+
2
t
=
2
∫
d
t
t
2
+
2
t
+
17
=
2
∫
d
t
(
t
+
1
)
2
+
16
=
{\displaystyle =\int {\frac {2{\mathsf {d}}t}{9(1+t^{2})+8(1-t^{2})+2t}}=2\int {\frac {{\mathsf {d}}t}{9+9t^{2}+8-8t^{2}+2t}}=2\int {\frac {{\mathsf {d}}t}{t^{2}+2t+17}}=2\int {\frac {{\mathsf {d}}t}{(t+1)^{2}+16}}=}
=
2
∫
d
t
16
(
(
t
+
1
)
2
16
+
1
)
=
2
∫
4
d
(
t
+
1
4
)
16
(
(
t
+
1
4
)
2
+
1
)
=
1
2
∫
d
(
t
+
1
4
)
(
t
+
1
4
)
2
+
1
=
1
2
arctan
t
+
1
4
+
C
=
1
2
arctan
tan
x
2
+
1
4
+
C
,
{\displaystyle =2\int {\frac {{\mathsf {d}}t}{16({\frac {(t+1)^{2}}{16}}+1)}}=2\int {\frac {4{\mathsf {d}}({\frac {t+1}{4}})}{16(({\frac {t+1}{4}})^{2}+1)}}={\frac {1}{2}}\int {\frac {{\mathsf {d}}({\frac {t+1}{4}})}{({\frac {t+1}{4}})^{2}+1}}={\frac {1}{2}}\arctan {\frac {t+1}{4}}+C={\frac {1}{2}}\arctan {\frac {\tan {\frac {x}{2}}+1}{4}}+C,}
kur
d
(
t
+
1
4
)
=
d
t
4
;
d
t
=
4
d
(
t
+
1
4
)
.
{\displaystyle {\mathsf {d}}\left({\frac {t+1}{4}}\right)={\frac {{\mathsf {d}}t}{4}};\;\;{\mathsf {d}}t=4{\mathsf {d}}\left({\frac {t+1}{4}}\right).}
∫
1
+
sin
x
sin
x
(
1
+
cos
x
)
d
x
=
∫
(
1
+
2
t
1
+
t
2
)
2
d
t
1
+
t
2
2
t
1
+
t
2
(
1
+
1
−
t
2
1
+
t
2
)
=
∫
1
+
t
2
+
2
t
1
+
t
2
2
d
t
1
+
t
2
2
t
1
+
t
2
(
1
+
t
2
1
+
t
2
+
1
−
t
2
1
+
t
2
)
=
∫
2
⋅
t
2
+
2
t
+
1
(
1
+
t
2
)
2
d
t
2
t
1
+
t
2
⋅
2
1
+
t
2
=
∫
2
(
t
2
+
2
t
+
1
)
d
t
4
t
=
1
2
∫
(
t
+
2
+
1
t
)
d
t
=
.
{\displaystyle \int {\frac {1+\sin x}{\sin x(1+\cos x)}}dx=\int {\frac {(1+{\frac {2t}{1+t^{2}}}){\frac {2\;{\mathsf {d}}t}{1+t^{2}}}}{{\frac {2t}{1+t^{2}}}(1+{\frac {1-t^{2}}{1+t^{2}}})}}=\int {\frac {{\frac {1+t^{2}+2t}{1+t^{2}}}{\frac {2\;{\mathsf {d}}t}{1+t^{2}}}}{{\frac {2t}{1+t^{2}}}({\frac {1+t^{2}}{1+t^{2}}}+{\frac {1-t^{2}}{1+t^{2}}})}}=\int {\frac {2\cdot {\frac {t^{2}+2t+1}{(1+t^{2})^{2}}}\;{\mathsf {d}}t}{{\frac {2t}{1+t^{2}}}\cdot {\frac {2}{1+t^{2}}}}}=\int {\frac {2(t^{2}+2t+1)\;{\mathsf {d}}t}{4t}}={\frac {1}{2}}\int (t+2+{\frac {1}{t}}){\mathsf {d}}t=.}
=
1
2
(
t
2
2
+
2
t
+
ln
t
)
+
C
=
t
2
4
+
t
+
1
2
ln
t
+
C
=
tan
2
x
2
4
+
tan
x
2
+
1
2
ln
tan
x
2
+
C
.
{\displaystyle ={\frac {1}{2}}({\frac {t^{2}}{2}}+2t+\ln t)+C={\frac {t^{2}}{4}}+t+{\frac {1}{2}}\ln t+C={\frac {\tan ^{2}{\frac {x}{2}}}{4}}+\tan {\frac {x}{2}}+{\frac {1}{2}}\ln \tan {\frac {x}{2}}+C.}
keisti
Jeigu pointegralinė funkcija turi pavidalą
R
(
sin
(
x
)
,
cos
(
x
)
)
{\displaystyle R(\sin(x),\cos(x))}
sin
x
{\displaystyle \sin x}
cos
x
{\displaystyle \cos x}
tan
x
=
t
,
{\displaystyle \tan x=t,}
x
=
arctan
t
,
{\displaystyle x=\arctan t,}
nes
sin
2
x
{\displaystyle \sin ^{2}x}
cos
2
x
{\displaystyle \cos ^{2}x}
tan
x
{\displaystyle \tan x}
cos
2
x
=
1
2
(
1
+
1
−
tan
2
x
1
+
tan
2
x
)
=
1
2
⋅
1
+
tan
2
x
+
1
−
tan
2
x
1
+
tan
2
x
=
1
2
⋅
2
1
+
tan
2
x
=
1
1
+
tan
2
x
=
1
1
+
t
2
,
{\displaystyle \cos ^{2}x={\frac {1}{2}}\left(1+{1-\tan ^{2}x \over 1+\tan ^{2}x}\right)={\frac {1}{2}}\cdot {1+\tan ^{2}x+1-\tan ^{2}x \over 1+\tan ^{2}x}={\frac {1}{2}}\cdot {2 \over 1+\tan ^{2}x}={1 \over 1+\tan ^{2}x}={1 \over 1+t^{2}},}
cos
2
x
=
cos
2
x
cos
2
x
+
sin
2
x
=
1
cos
2
x
+
sin
2
x
cos
2
x
=
1
1
+
tan
2
x
=
1
1
+
t
2
;
{\displaystyle \cos ^{2}x={\frac {\cos ^{2}x}{\cos ^{2}x+\sin ^{2}x}}={\frac {1}{\frac {\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}}}={1 \over 1+\tan ^{2}x}={1 \over 1+t^{2}};}
sin
2
x
=
1
2
(
1
−
1
−
tan
2
x
1
+
tan
2
x
)
=
1
2
⋅
1
+
tan
2
x
−
(
1
−
tan
2
x
)
1
+
tan
2
x
=
1
2
⋅
2
tan
2
x
1
+
tan
2
x
=
tan
2
x
1
+
tan
2
x
=
t
2
1
+
t
2
,
{\displaystyle \sin ^{2}x={\frac {1}{2}}\left(1-{1-\tan ^{2}x \over 1+\tan ^{2}x}\right)={\frac {1}{2}}\cdot {1+\tan ^{2}x-(1-\tan ^{2}x) \over 1+\tan ^{2}x}={\frac {1}{2}}\cdot {2\tan ^{2}x \over 1+\tan ^{2}x}={\tan ^{2}x \over 1+\tan ^{2}x}={t^{2} \over 1+t^{2}},}
sin
2
x
=
sin
2
x
cos
2
x
+
sin
2
x
=
sin
2
x
cos
2
x
cos
2
x
+
sin
2
x
cos
2
x
=
tan
2
x
1
+
tan
2
x
=
t
2
1
+
t
2
;
{\displaystyle \sin ^{2}x={\frac {\sin ^{2}x}{\cos ^{2}x+\sin ^{2}x}}={\frac {\frac {\sin ^{2}x}{\cos ^{2}x}}{\frac {\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}}}={\tan ^{2}x \over 1+\tan ^{2}x}={t^{2} \over 1+t^{2}};}
d
x
=
d
(
arctan
t
)
=
d
t
1
+
t
2
.
{\displaystyle {\mathsf {d}}x={\mathsf {d}}(\arctan t)={\frac {{\mathsf {d}}t}{1+t^{2}}}.}
∫
d
x
2
−
sin
2
x
=
∫
d
t
(
2
−
t
2
1
+
t
2
)
(
1
+
t
2
)
=
∫
d
t
(
2
(
1
+
t
2
)
−
t
2
1
+
t
2
)
(
1
+
t
2
)
=
∫
d
t
(
2
+
2
t
2
−
t
2
1
+
t
2
)
(
1
+
t
2
)
=
{\displaystyle \int {\frac {{\mathsf {d}}x}{2-\sin ^{2}x}}=\int {\frac {{\mathsf {d}}t}{\left(2-{t^{2} \over 1+t^{2}}\right)(1+t^{2})}}=\int {\frac {{\mathsf {d}}t}{\left({2(1+t^{2})-t^{2} \over 1+t^{2}}\right)(1+t^{2})}}=\int {\frac {{\mathsf {d}}t}{\left({2+2t^{2}-t^{2} \over 1+t^{2}}\right)(1+t^{2})}}=}
=
∫
d
t
2
+
t
2
=
∫
1
2
arctan
t
2
+
C
=
1
2
arctan
(
tan
x
2
)
+
C
.
{\displaystyle =\int {\frac {{\mathsf {d}}t}{2+t^{2}}}=\int {\frac {1}{\sqrt {2}}}\arctan {\frac {t}{\sqrt {2}}}+C={\frac {1}{\sqrt {2}}}\arctan \left({\frac {\tan x}{\sqrt {2}}}\right)+C.}
Patikriname
d
d
x
(
1
2
arctan
(
tan
x
2
)
+
C
)
=
1
2
d
d
x
(
arctan
(
tan
x
2
)
)
.
{\displaystyle {\frac {\mathsf {d}}{{\mathsf {d}}x}}\left({\frac {1}{\sqrt {2}}}\arctan \left({\frac {\tan x}{\sqrt {2}}}\right)+C\right)={\frac {1}{\sqrt {2}}}{\frac {\mathsf {d}}{{\mathsf {d}}x}}\left(\arctan \left({\frac {\tan x}{\sqrt {2}}}\right)\right).}
Turime, kad
d
d
u
arctan
u
=
1
1
+
u
2
,
{\displaystyle {\frac {\mathsf {d}}{{\mathsf {d}}u}}\arctan u={\frac {1}{1+u^{2}}},}
u
=
tan
x
2
{\displaystyle u={\frac {\tan x}{\sqrt {2}}}}
d
d
x
(
tan
x
2
)
=
1
2
cos
2
x
.
{\displaystyle {\frac {\mathsf {d}}{{\mathsf {d}}x}}({\frac {\tan x}{\sqrt {2}}})={\frac {1}{{\sqrt {2}}\cos ^{2}x}}.}
1
2
d
d
u
(
arctan
u
)
d
d
x
(
tan
x
)
=
1
2
1
1
+
u
2
1
2
cos
2
x
=
1
2
1
1
+
(
tan
x
2
)
2
1
2
cos
2
x
=
1
2
1
1
+
tan
2
x
2
1
2
cos
2
x
=
{\displaystyle {\frac {1}{\sqrt {2}}}{\frac {\mathsf {d}}{{\mathsf {d}}u}}\left(\arctan u\right){\frac {\mathsf {d}}{{\mathsf {d}}x}}\left(\tan x\right)={\frac {1}{\sqrt {2}}}{\frac {1}{1+u^{2}}}{\frac {1}{{\sqrt {2}}\cos ^{2}x}}={\frac {1}{\sqrt {2}}}{\frac {1}{1+({\frac {\tan x}{\sqrt {2}}})^{2}}}{\frac {1}{{\sqrt {2}}\cos ^{2}x}}={\frac {1}{\sqrt {2}}}{\frac {1}{1+{\frac {\tan ^{2}x}{2}}}}{\frac {1}{{\sqrt {2}}\cos ^{2}x}}=}
=
1
2
2
2
+
tan
2
x
1
cos
2
x
=
1
2
2
2
+
sin
2
x
cos
2
x
1
cos
2
x
=
1
2
2
2
cos
2
x
+
sin
2
x
cos
2
x
1
cos
2
x
=
1
2
2
cos
2
x
2
cos
2
x
+
sin
2
x
1
cos
2
x
=
1
2
2
2
cos
2
x
+
sin
2
x
=
{\displaystyle ={\frac {1}{2}}{\frac {2}{2+\tan ^{2}x}}{\frac {1}{\cos ^{2}x}}={\frac {1}{2}}{\frac {2}{2+{\frac {\sin ^{2}x}{\cos ^{2}x}}}}{\frac {1}{\cos ^{2}x}}={\frac {1}{2}}{\frac {2}{\frac {2\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}}}{\frac {1}{\cos ^{2}x}}={\frac {1}{2}}{\frac {2\cos ^{2}x}{2\cos ^{2}x+\sin ^{2}x}}{\frac {1}{\cos ^{2}x}}={\frac {1}{2}}{\frac {2}{2\cos ^{2}x+\sin ^{2}x}}=}
=
1
2
(
1
−
sin
2
x
)
+
sin
2
x
=
1
2
−
2
sin
2
x
+
sin
2
x
=
1
2
−
sin
2
x
.
{\displaystyle ={\frac {1}{2(1-\sin ^{2}x)+\sin ^{2}x}}={\frac {1}{2-2\sin ^{2}x+\sin ^{2}x}}={\frac {1}{2-\sin ^{2}x}}.}
