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}}}.}
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.}
∫
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;}
cos
x
=
1
1
+
t
2
;
{\displaystyle \cos x={\frac {1}{\sqrt {1+t^{2}}}};}
1
cos
2
x
=
1
(
1
1
+
t
2
)
2
=
1
+
t
2
;
{\displaystyle {\frac {1}{\cos ^{2}x}}={\frac {1}{({\frac {1}{\sqrt {1+t^{2}}}})^{2}}}=1+t^{2};}
d
x
cos
2
x
=
d
t
1
+
t
2
(
1
1
+
t
2
)
2
=
d
t
.
{\displaystyle {\frac {dx}{\cos ^{2}x}}={\frac {\frac {dt}{1+t^{2}}}{({\frac {1}{\sqrt {1+t^{2}}}})^{2}}}=dt.}
∫
sin
2
x
cos
6
x
d
x
=
∫
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 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.}
∫
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
=
−
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}}}=-{\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
=
9
∫
sin
t
3
9
−
9
sin
2
t
3
⋅
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=9\int \sin t{\frac {3{\sqrt {9-9\sin ^{2}t}}}{3}}\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
tan
2
t
=
1
a
∫
d
t
sin
t
=
1
a
ln
|
1
−
cos
t
sin
t
|
+
C
=
{\displaystyle \int {\frac {dx}{x{\sqrt {a^{2}+x^{2}}}}}=\int {\frac {a\;dt}{a\cos ^{2}t\cdot \tan t{\sqrt {a^{2}\tan ^{2}t}}}}={\frac {1}{a}}\int {\frac {dt}{\sin t}}={\frac {1}{a}}\ln |{\frac {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}{\sin t}}-\cot t|+C={\frac {1}{a}}\ln |{\frac {{\sqrt {a^{2}+x^{2}}}-a}{x}}|+C,}
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}}.}
