∫R(sinx,cosx
[Pages:8]Integrals-tasks (VII part) Integration of some trigonometric functions
We'll give you tips for four types of integrals of trigonometric functions.
A) Integral type R(sin x, cos x)dx
These are integrals in which the sinx and cosx do not have degrees . Here we introduce substitution tg x = t 2
From substitution we get : (using the formula from trigonometry)
x
sin x =
sin x 1
2sin x cos x
=
22
sin2 x + cos2 x
2
2
=
cos2
x 2
2
sin 2
cos x 2
cos2
x 2
sin 2 cos2
x
2 x
2
+1
2tg x = 2=
tg 2 x +1 2
2t = 2t t2 +1 1+t2
cos x =
cos x 1
=
cos2 x - sin2 2
sin2 x + cos2 2
x
2 x
2
=
cos2 cos2
x 2
x 2
1
-
sin 2 cos2
x
2 x
2
sin 2 cos2
x
2 x
2
+1
1- tg 2 =
1+ tg 2
x
2 x
2
= 1-t2 1+ t2
As tg x = t then 2
x 2
=
arctgt
x
=
2arctgt
dx
=
2 1+ t2
dt
To summarize:
When we take substitution tg x = t we have: 2
sin
x
=
1
2t +t
2
cos
x
=
1 1
- +
t t
2 2
dx
=
2 1+ t2
dt
Substitution tg x = t is universal trigonometric substitution and can always be used, but it's easier, depending on the 2
look of function to use the following substitution:
1
B) Integral type R(tgx)dx and R(sin2x, cos2 x,sin x cos x)dx
These are integrals where the function under the integral can be reduced to tgx or that are occurring degree of sinus and cosine and product sin x cos x
Here we introduce substitution : tgx = t From substitution we get : (using the formula from trigonometry)
sin2 x
sin 2
x=
sin 2 1
x
=
sin2 x sin2 x + cos2
x
=
cos2 x sin2 x + cos2
x
=
tg 2 x = t 2 tg 2 x +1 1+ t2
cos2 x cos2 x
cos2 x
cos2
x=
cos2 1
x
=
cos2 x sin2 x + cos2
x
=
cos2 x sin2 x + cos2
x
=
1 =1 tg 2 x +1 1+ t2
cos2 x cos2 x
sin x cos x
sin x cos x =
sin x cos x 1
=
sin x cos x sin2 x + cos2 x
=
cos 2 x sin2 x + cos2
x
=
tgx = t tg 2x +1 1+ t2
cos2 x cos2 x
tg
x 2
=
t
x 2
=
arctgt
x
=
2arctgt
dx
=
2 1+ t2
dt
To summarize:
When we take substitution tgx = t we have :
sin
2
x
=
1
t2 +t
2
cos2
x
=
1 1+ t2
sin
x
cos
x
=
1
t +
t
2
dx
=
2 1+ t2
dt
2
C) Integral type sinm x cosn xdx
Two situations: i) If m and n are integers ii) If m and n are rational numbers In both situations we introduce substitution sin x = u or cosx=u but, i) when m and n are integers integral is reduced to integration of rational function ii) when m and n are rational numbers we have integral of differential binomial
D) Integral type sin ax cos bxdx; sin ax sin bxdx; cos ax cos bxdx;
First, use the trigonometric formulas: sinax sinbx= 1 [cos(a-b)x ? cos(a+b)x] 2 sinax cosbx= 1 [sin(a+b)x + sin(a-b)x] 2 cosax cosbx= 1 [cos(a+b)x + cos(a-b)x] 2
Some "tricks": If in integral we have expression a 2 - x2 , it is convenient to take x = asint as it "destroyed " the root:
a 2 - x 2 = a2 - (a sin t)2 = a2 - a2 (sin t)2 =a 1- sin2 t =a cos t
If in integral we have expression x2 + a 2 , it is convenient to take x=a tgt as it "destroyed " the root:
x 2 + a 2 = (atgt)2 + a2 = a2tg 2t + a2 =a tg 2t +1 = a sin2 t +1 = a sin2 t + cos2 t = a 1 = a 1
cos2 t
cos2 t
cos2 t cos t
3
Examples
Example 1.
dx = ? sin x
This integral we have already solved in Integrals-tasks (I part) without using trigonometric substitution.
Now, we can use
tg x = t .
So :
sin
x
=
1
2t +t
2
2
dx
=
2 1+ t2
dt
2
dx sin x
=
1+ t2 2t
dt
=
1dt t
=
ln
t
+C
=
ln
tg
x 2
+C
1+ t2
Example 2.
2- 2+
sin cos
x dx x
=
?
Here we use tg x = t because sinx and cosx have no degrees... 2
tg x = t 2
dx
=
2 1+ t2
dt
sin
x
=
1
2t +t
2
cos
t
=
1- 1+
t t
2 2
2 2
- +
sin cos
x x
dx
=
2
-
1
2t +t
2
2
+
1 1
- +
t t
2 2
2 1+ t2
dt
=
2 + 2t2 - 2t
1+ t2
2
2 + 2t2 +1- t 2 1+ t 2
dt = 4
t2 -t +1 (t 2 + 3)(1+ t 2 ) dt
1+ t2
This is integral of rational function ...
4
Beware: both terms in the denominator are irresolvable:
So:
t2 -t +1 (t2 + 3)(1+ t2 )
=
At t2
+B +3
+
Ct + 1+ t
D
2
.....................................
/
(t
2
+ 3)(1+ t2 )
t2 - t +1 = At + At3 + B + Bt2 + Ct3 + 3Ct + Dt2 + 3D
t2 - t +1 ( A + C)t3 + (B + D)t2 + ( A + 3C)t + B + 3D
A + C = 0, B + D = 1, A + 3C = -1 i B + 3D = 1 A = 1/ 2, B = 1, C = -1/ 2, D = 0
4
t2 -t +1 (t2 + 3)(1+ t2 ) dt = 4
1 t +1
2 t2
+
3
dt
+
4
-1 t
2 1+
t2
dt
=
2
t+2 t2 +3
dt
-
2
t 1+ t2 dt
Solution is:
=
ln
t2 + 1+ t
3
2
+
2 arctg 3
t
3 + tg 2 x
+ C = ln
2+
3
1+ tg 2 x
2
tg x arctg 2 + C
3
3
2
5
Example 3.
I
=
(2
+
dx cos x) sin
x
=
?
Often integral to give a letter mark, most frequently used letters : I, J ...
Sabstitution: tg x = t . 2
tg x = t 2
dx
=
2 1+ t2
dt
sin
x
=
2t 1+ t
2
cos t
=
1- 1+
t2 t2
2
So:
I =
1+ t2
2
+
1- 1+
t2 t2
2t 1+ t2
dt
=
dt
=
2
+
2t 2 +1- t 2 1+ t2
t
1+ t2 (t2 + 3)t dt
Now, we do:
1+ t2 = A + Bt + C = At2 + 3A + Bt2 + Ct2
(t2 + 3)t t t2 + 3
t(t2 + 3)
1+ t2 = A(t2 + 3) + (Bt + C)t
1+ t2 = At2 + 3A + Bt2 + Ct
1+ t2 = t2 ( A + B) + Ct + 3A
A + B = 1, C = 0, 3A = 1
A = 1/ 3, B = 2 / 3, C = 0
t2 +3 = u
I
=
1 3
dt t
+
2 3
t2
t +
3
dt
=
tdt
=
1 du
=
1 ln 3
t
+1 3
du u
+c
2
I
= 1 ln 3
t
+ 1 ln(t2 + 3) + C = 3
1 ln tg x 32
+
1 3
ln
tg
2
x 2
+
3
+
C
6
Example 4.
I=
sin x cos x sin4 x + cos4
x
dx
=
?
tgx = t
sin2
x
=
t2 1+ t2
So:
cos2
x
=
1 1+ t2
sin
x
cos
x
=
1
t +
t
2
dx
=
2 1+ t2
dt
t
t
I =
sin x cos x sin4 x + cos4
x
dx
=
1
t2 +t
2
1+ t2
2
+
1 1+ t2
2
2 1+ t2
dt
=
1+ t2 t4 +1
2 dt = 1+ t2
(1+ t2 ) 2
=
2t
t4
+
dt 1
=
2t
(t 2
)2
+
dt 1
t2 = z 2tdt = dz
=
dz z2 +1
=
arctgz
+
C
=
arctgt 2
+
C
=
arctg(tg 2
x)
+
C
This task we could solve in another way, using trigonometric formulas
The idea is to transform the expression in the denominator. Start from the basic identity:
sin2 x + cos2 x = 1............................. / ()2 sin4 x + 2 sin2 x cos2 x + cos4 x = 1 sin4 x + cos4 x = 1- 2 sin2 x cos2 x sin4 x + cos4 x = 1- 4 sin2 x cos2 x
2 sin4 x + cos4 x = 1- sin2 2x
2 sin4 x + cos4 x = 2 - sin2 2x
2 sin4 x + cos4 x = 1+ 1- sin2 2x
2 sin4 x + cos4 x = 1+ cos2 2x
2
7
Return to the integral:
cos 2x = t
I
=
sin x cos x 1+ cos2 2x
dx
=
2 sin x cos x 1+ cos2 2x dx
=
sin 2x
1
+
cos
2
2
dx x
- sin 2x 2dx = dt
=
1 -2
1
dt +t
2
2
sin
2xdx
=
dt -2
=
1 -2
arctgt
+
C
=
-1 2
arctg(cos 2x) + C
Example 5.
a2 - x2 = ?
Remember this integral? We solved it so far in two ways: a partial integration and using Ostrogradski formula... Here's another way of solving:
If we use sabstitution x = a sin t
a 2 - x 2 = a2 - (a sin t)2 = a2 - a2 (sin t)2 =a 1- sin2 t =a cos t
x = a sin t dx = a cos tdt
a2 - x2 = a cos t a cos tdt = a2 cos2tdt = a2 1+ cos 2tdt =
2
a2 (1+ cos 2t)dt = a2 (t + 1 sin 2t) + C =
2
22
From substitution , we have:
x = a sin t sin t = x t = arcsin x go back...
a
a
1 sin 2t = 1 2 sin t cos t = sin t 1- sin2 t = sin(arcsin x ) 1- sin2 (arcsin x ) =
2
2
a
a
=x a
1-
x2 a2
=
x a
a2 - x2 = x
a2
a2
a2 - x2
Solution is:
a2 2
(t
+
1 2
sin
2t )
+
C
=
a2 2
(arcsin
x a
+
x a2
a2 - x2 ) + C = a2 arcsin x + x
2
a2
a2 - x2 + C
8
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