Trigonometric Substitutions Sometimes it can be beneficial to switch out the original variable for a more complex one. Although it appears to be a “reverse” substitution, it actually works the same way as a regular replacement in theory.
When the function you want to integrate has a polynomial expression that would allow you to employ the fundamental identity, this form of substitution is typically suggested
sin^2x+cos^2x = 1sin^2x+cos^2x = 1 one of three ways:
Cos^2x = 1โsin^2x sec^2x = 1+tan^2x tan^2x = sec^2xโ1.
Find Out Integrals Using Trigonometric Substitutions
There are two techniques to solve Trig substitution, most quick method is to use trig sub calculator and get the answer without any difficult calculation. The second method is the manual calculation and understanding the formula and calculation, so read the article for understanding the trigonometric substitution.
The Trigonometric Substitutions technique is used to solve the following integration issues. It is a technique for determining the ant derivatives of functions that contain rational powers of quadratic expressions of the form n2 (where n is an integer) or square roots of quadratic expressions. Such expressions include, for instance,
4โx2โ and (x2+1)3/24โx2 and (x2+1)3/2
When more popular and user-friendly methods of integration have failed, the trig substitution approach may be used. Trigonometric substitution relies on the knowledge of standard trigonometric denials, differential notation, u-substitution integration, and integration of trigonometric functions.
Remember that if
x = f( ฮธ)ย ,x = f( ฮธ)ย ,
dx = fโฒ( ฮธ)ย d ฮธdx = fโฒ( ฮธ)ย d ฮธ
example, if
x = sec ฮธ ,x = sec ฮธ ,
Then
dx = sec ฮธtan ฮธย d ฮธdx = sec ฮธtan ฮธย d ฮธ
The purpose of trig substitution is to replace integer powers of trig functions with square roots of quadratic expressions, which may be difficult to integrate using other methods of integration, or rational powers of quadratic expressions of the form n2 n2 (where n n is an integer). For instance, if we begin with the phrase
4โx2โ4โx2
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Letโs suppose
x = 2sin ฮธ ,x = 2sin ฮธ,
then
4โx2โ = 4โ(2sin ฮธ)2โ4โx2 = 4โ(2sin ฮธ)2
= 4โ4sin2 ฮธโ = 4โ4sin2 ฮธ
= 4(1โsin2 ฮธ)โ = 4(1โsin2 ฮธ)
= 4โโโ
1โsin2 ฮธโโโโ = 4โ
1โsin2 ฮธ
(Recall thatย ย cos2 ฮธ+sin2 ฮธ = 1ย ย cos2 ฮธ+sin2 ฮธ = 1 so thatย ย 1โsin2 ฮธ = cos2 ฮธย 1โsin2 ฮธ = cos2 ฮธ.)
= 2โ
cos2 ฮธโ = 2โ
cos2 ฮธ
= 2โ
โฃโฃcos ฮธโฃโฃ = 2โ
|cos ฮธ|
(Assume thatย ย โฯ2โค ฮธโคฯ2ย ย โฯ2โค ฮธโคฯ2 so thatย ย cos ฮธโฅ0 cos ฮธโฅ0.)
= 2cos ฮธ = 2cos ฮธ
and
dx = 2cos ฮธย d ฮธdx = 2cos ฮธย d ฮธ
Thus,
โซ4โx2โdxโซ4โx2dx
could be rewritten as
โซ4โx2โdx = โซ2cos ฮธโ
2cos ฮธd ฮธ = 4โซcos2 ฮธd ฮธโซ4โx2dx = โซ2cos ฮธโ
2cos ฮธd ฮธ = 4โซcos2 ฮธd ฮธ
(Recall thatย ย cos2 ฮธ = 2cos2 ฮธโ1ย ย cos2 ฮธ = 2cos2 ฮธโ1 so thatย ย cos2 ฮธ = 12(1+cos2 ฮธ)ย cos2 ฮธ = 12(1+cos2 ฮธ)ย .)
= 4โซ12(1+cos2 ฮธ)d ฮธ = 4โซ12(1+cos2 ฮธ)d ฮธ
= 2โซ(1+cos2 ฮธ)d ฮธ = 2โซ(1+cos2 ฮธ)d ฮธ
= 2( ฮธ+12sin2 ฮธ)+C = 2( ฮธ+12sin2 ฮธ)+C
= 2 ฮธ+sin2 ฮธ+C = 2 ฮธ+sin2 ฮธ+C
(Recall thatย ย sin2 ฮธ = 2sin ฮธcos ฮธย sin2 ฮธ = 2sin ฮธcos ฮธย .)
= 2 ฮธ+2sin ฮธcos ฮธ+C = 2 ฮธ+2sin ฮธcos ฮธ+C
Our final response must be expressed in terms of x. Given that x = 2sin x, it follows that
sin ฮธ = x2 = opposite hypotenuse sin ฮธ = x2 = opposite hypotenuse
and
ฮธ = arcsin(x2) ฮธ = arcsin(x2)
Using the given right triangle and the Pythagorean Theorem, we can determine any trig value of ฮธ.
Since
(adjacent)2+(opposite)2 = (hypotenuse)2 โถ(adjacent)2+(opposite)2 = (hypotenuse)2 โถ
(adjacent)2+(x)2 = (2)2 โถ adjacent = 4โx2โ โถ(adjacent)2+(x)2 = (2)2 โถ adjacent = 4โx2 โถ
Then
2 ฮธ+2sin ฮธcos ฮธ+C = 2arcsin(x2)+2โ
x2โ
4โx2โ22 ฮธ+2sin ฮธcos ฮธ+C = 2arcsin(x2)+2โ
x2โ
4โx22
= 2arcsin(x2)+12xโ
4โx2โ+C = 2arcsin(x2)+12xโ
4โx2+C
When use the trig substitution method, we always follow 3 popular trig identities:
(I)ย 1โsin2 ฮธ = cos2 ฮธ 1โsin2 ฮธ = cos2 ฮธ
(II)ย 1+tan2 ฮธ = sec2 ฮธ 1+tan2 ฮธ = sec2 ฮธย and
(III)ย sec^2 ฮธโ1 = tan2 ฮธ sec2 ฮธโ1 = tan2 ฮธ
For the expression
a2โx2โโa2โx2
we use equation (I) and let
x = asin ฮธx = asin ฮธ
(Assume that 2 is 2 is 2 is 2 and that cos 0 is cos 0). This permits x to have both positive and negative values.)
Then
a2โx2โโ = a2โa2sin2 ฮธโโa2โx2 = a2โa2sin2 ฮธ
= a2(1โsin2 ฮธ)โโ = a2(1โsin2 ฮธ)
= a2cos2 ฮธโโโ = a2cos2 ฮธ
= a2โโโcos2 ฮธโ = a2cos2 ฮธ
= acos2 ฮธโ = acos2 ฮธ
= aโฃโฃcos ฮธโฃโฃ = a|cos ฮธ|
= acos ฮธ = acos ฮธ
and
dx = acos ฮธย d ฮธdx = acos ฮธย d ฮธ
For the expression
a2+x2โโa2+x2
we use equation (II) and let
x = atan ฮธx = atan ฮธ
(Presume that 2 is 2 is 2 is 2 and that cos > 0 is 0 and sec > 0 is 0). This permits x x to have both positive and negative values.)
Then
a2+x2โโ = a2+a2tan2 ฮธโโa2+x2 = a2+a2tan2 ฮธ
= a2(1+tan2 ฮธ)โโ = a2(1+tan2 ฮธ)
= a2โโโsec2 ฮธโ = a2sec2 ฮธ
= asec2 ฮธโ = asec2 ฮธ
= aโฃโฃsec ฮธโฃโฃ = a|sec ฮธ|
= asec ฮธ = asec ฮธ
and
dx = asec2 ฮธย d ฮธdx = asec2 ฮธย d ฮธ
For the expression
x2โa2โโx2โa2
we use equation (3) and let
x = asec ฮธx = asec ฮธ
(Assume that 0 is 2, 0 is 2, and that tan 0 is tan 0). This limits the values of x to only positive numbers. You must use 2 with tan2 = tan tan2 = tan if the integral contains negative values of x x.)
Then
x2โa2โโ = a2sec2 ฮธโa2โโx2โa2 = a2sec2 ฮธโa2
= a2(sec2 ฮธโ1)โโ = a2(sec2 ฮธโ1)
= a2โโโsec2 ฮธโ1โโโโ = a2sec2 ฮธโ1
= atan2 ฮธโ = atan2 ฮธ
= aโฃโฃtan ฮธโฃโฃ = a|tan ฮธ|
= atan ฮธ = atan ฮธ
and
dx = asec ฮธtan ฮธย d ฮธdx = asec ฮธtan ฮธย d ฮธ
Basic indefinite trigonometric integral formulas :
- โซcosxย dx = sinx+C โซcosxย dx = sinx+C
- โซsinxย dx = โcosx+C โซsinxย dx = โcosx+C
- โซsec2xย dx = tanx+C โซsec2xย dx = tanx+C
- โซcsc2xย dx = โcotx+C โซcsc2xย dx = โcotx+C
- โซsecxtanxย dx = secx+C โซsecxtanxย dx = secx+C
- โซcscxcotxย dx = โcscx+C โซcscxcotxย dx = โcscx+C
- โซtanxย dx = ln|secx|+C โซtanxย dx = ln|secx|+C
- โซcotxย dx = ln|sinx|+C โซcotxย dx = ln|sinx|+C
- โซsecxย dx = ln|secx+tanx|+C โซsecxย dx = ln|secx+tanx|+C
- โซcscxย dx = ln|cscxโcotx|+C โซcscxย dx = ln|cscxโcotx|+C
The majority of the issues listed below are typical. A handful of them pose some difficulties. Use the differential notation dx and d precisely, and use caution when simplifying expressions algebraically and arithmetically. You should be capable to rationally integrating a variety of powers and trig functions. The following additional well-known trig identities may be required.
- sin2x = 2sinxcosx sin2x = 2sinxcosx
- cos2x = 2cos2xโ1 cos2x = 2cos2xโ1 ย so thatย cos2x = 12(1+cos2x) cos2x = 12(1+cos2x)
- cos2x = 1โ2sin2x cos2x = 1โ2sin2x ย so thatย sin2x = 12(1โcos2x) sin2x = 12(1โcos2x)
- cos2x = cos2xโsin2x cos2x = cos2xโsin2x
- 1+cot2x = csc2x 1+cot2x = csc2x ย so thatย cot2x = csc2xโ1 cot2x = csc2xโ1