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 = 1âˆ’sin^2x sec^2x = 1+tan^2x tan^2x = sec^2xâˆ’1.

## 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,

4âˆ’x2âˆš and (x2+1)3/24âˆ’x2 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( Î¸)Â ,x = f( Î¸)Â ,

dx = fâ€²( Î¸)Â d Î¸dx = fâ€²( Î¸)Â d Î¸

example, if

x = sec Î¸ ,x = sec Î¸ ,

Then

dx = sec Î¸tan Î¸Â d Î¸dx = sec Î¸tan Î¸Â d Î¸

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

4âˆ’x2âˆš4âˆ’x2

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**Letâ€™s suppose**

x = 2sin Î¸ ,x = 2sin Î¸,

then

4âˆ’x2âˆš = 4âˆ’(2sin Î¸)2âˆš4âˆ’x2 = 4âˆ’(2sin Î¸)2

= 4âˆ’4sin2 Î¸âˆš = 4âˆ’4sin2 Î¸

= 4(1âˆ’sin2 Î¸)âˆš = 4(1âˆ’sin2 Î¸)

= 4â€“âˆšâ‹…1âˆ’sin2 Î¸âˆ’âˆ’âˆ’âˆš = 4â‹…1âˆ’sin2 Î¸

(Recall thatÂ Â cos2 Î¸+sin2 Î¸ = 1Â Â cos2 Î¸+sin2 Î¸ = 1 so thatÂ Â 1âˆ’sin2 Î¸ = cos2 Î¸Â 1âˆ’sin2 Î¸ = cos2 Î¸.)

= 2â‹…cos2 Î¸âˆš = 2â‹…cos2 Î¸

= 2â‹…âˆ£âˆ£cos Î¸âˆ£âˆ£ = 2â‹…|cos Î¸|

(Assume thatÂ Â âˆ’Ï€2â‰¤ Î¸â‰¤Ï€2Â Â âˆ’Ï€2â‰¤ Î¸â‰¤Ï€2 so thatÂ Â cos Î¸â‰¥0 cos Î¸â‰¥0.)

= 2cos Î¸ = 2cos Î¸

and

dx = 2cos Î¸Â d Î¸dx = 2cos Î¸Â d Î¸

Thus,

âˆ«4âˆ’x2âˆšdxâˆ«4âˆ’x2dx

could be rewritten as

âˆ«4âˆ’x2âˆšdx = âˆ«2cos Î¸â‹…2cos Î¸d Î¸ = 4âˆ«cos2 Î¸d Î¸âˆ«4âˆ’x2dx = âˆ«2cos Î¸â‹…2cos Î¸d Î¸ = 4âˆ«cos2 Î¸d Î¸

(Recall thatÂ Â cos2 Î¸ = 2cos2 Î¸âˆ’1Â Â cos2 Î¸ = 2cos2 Î¸âˆ’1 so thatÂ Â cos2 Î¸ = 12(1+cos2 Î¸)Â cos2 Î¸ = 12(1+cos2 Î¸)Â .)

= 4âˆ«12(1+cos2 Î¸)d Î¸ = 4âˆ«12(1+cos2 Î¸)d Î¸

= 2âˆ«(1+cos2 Î¸)d Î¸ = 2âˆ«(1+cos2 Î¸)d Î¸

= 2( Î¸+12sin2 Î¸)+C = 2( Î¸+12sin2 Î¸)+C

= 2 Î¸+sin2 Î¸+C = 2 Î¸+sin2 Î¸+C

(Recall thatÂ Â sin2 Î¸ = 2sin Î¸cos Î¸Â sin2 Î¸ = 2sin Î¸cos Î¸Â .)

= 2 Î¸+2sin Î¸cos Î¸+C = 2 Î¸+2sin Î¸cos Î¸+C

Our final response must be expressed in terms of x. Given that x = 2sin x, it follows that

sin Î¸ = x2 = opposite hypotenuse sin Î¸ = x2 = opposite hypotenuse

and

Î¸ = arcsin(x2) Î¸ = arcsin(x2)

Using the given right triangle and the Pythagorean Theorem, we can determine any trig value of Î¸.

Since

(adjacent)2+(opposite)2 = (hypotenuse)2 âŸ¶(adjacent)2+(opposite)2 = (hypotenuse)2 âŸ¶

(adjacent)2+(x)2 = (2)2 âŸ¶ adjacent = 4âˆ’x2âˆš âŸ¶(adjacent)2+(x)2 = (2)2 âŸ¶ adjacent = 4âˆ’x2 âŸ¶

Then

2 Î¸+2sin Î¸cos Î¸+C = 2arcsin(x2)+2â‹…x2â‹…4âˆ’x2âˆš22 Î¸+2sin Î¸cos Î¸+C = 2arcsin(x2)+2â‹…x2â‹…4âˆ’x22

= 2arcsin(x2)+12xâ‹…4âˆ’x2âˆš+C = 2arcsin(x2)+12xâ‹…4âˆ’x2+C

When use the trig substitution method, we always follow 3 popular trig identities:

(I)Â 1âˆ’sin2 Î¸ = cos2 Î¸ 1âˆ’sin2 Î¸ = cos2 Î¸

(II)Â 1+tan2 Î¸ = sec2 Î¸ 1+tan2 Î¸ = sec2 Î¸Â and

(III)Â sec^2 Î¸âˆ’1 = tan2 Î¸ sec2 Î¸âˆ’1 = tan2 Î¸

### For the expression

a2âˆ’x2âˆ’âˆša2âˆ’x2

we use equation (I) and let

x = asin Î¸x = asin Î¸

(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

a2âˆ’x2âˆ’âˆš = a2âˆ’a2sin2 Î¸âˆ’âˆša2âˆ’x2 = a2âˆ’a2sin2 Î¸

= a2(1âˆ’sin2 Î¸)âˆ’âˆš = a2(1âˆ’sin2 Î¸)

= a2cos2 Î¸âˆ’âˆ’âˆš = a2cos2 Î¸

= a2âˆ’âˆ’âˆšcos2 Î¸âˆš = a2cos2 Î¸

= acos2 Î¸âˆš = acos2 Î¸

= aâˆ£âˆ£cos Î¸âˆ£âˆ£ = a|cos Î¸|

= acos Î¸ = acos Î¸

and

dx = acos Î¸Â d Î¸dx = acos Î¸Â d Î¸

For the expression

a2+x2âˆ’âˆša2+x2

we use equation (II) and let

x = atan Î¸x = atan Î¸

(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+x2âˆ’âˆš = a2+a2tan2 Î¸âˆ’âˆša2+x2 = a2+a2tan2 Î¸

= a2(1+tan2 Î¸)âˆ’âˆš = a2(1+tan2 Î¸)

= a2âˆ’âˆ’âˆšsec2 Î¸âˆš = a2sec2 Î¸

= asec2 Î¸âˆš = asec2 Î¸

= aâˆ£âˆ£sec Î¸âˆ£âˆ£ = a|sec Î¸|

= asec Î¸ = asec Î¸

and

dx = asec2 Î¸Â d Î¸dx = asec2 Î¸Â d Î¸

For the expression

x2âˆ’a2âˆ’âˆšx2âˆ’a2

we use equation (3) and let

x = asec Î¸x = asec Î¸

(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

x2âˆ’a2âˆ’âˆš = a2sec2 Î¸âˆ’a2âˆ’âˆšx2âˆ’a2 = a2sec2 Î¸âˆ’a2

= a2(sec2 Î¸âˆ’1)âˆ’âˆš = a2(sec2 Î¸âˆ’1)

= a2âˆ’âˆ’âˆšsec2 Î¸âˆ’1âˆ’âˆ’âˆ’âˆš = a2sec2 Î¸âˆ’1

= atan2 Î¸âˆš = atan2 Î¸

= aâˆ£âˆ£tan Î¸âˆ£âˆ£ = a|tan Î¸|

= atan Î¸ = atan Î¸

and

dx = asec Î¸tan Î¸Â d Î¸dx = asec Î¸tan Î¸Â d Î¸

### Basic indefinite trigonometric integral formulas :

- âˆ«cosxÂ dx = sinx+C âˆ«cosxÂ dx = sinx+C
- âˆ«sinxÂ dx = âˆ’cosx+C âˆ«sinxÂ dx = âˆ’cosx+C
- âˆ«sec2xÂ dx = tanx+C âˆ«sec2xÂ dx = tanx+C
- âˆ«csc2xÂ dx = âˆ’cotx+C âˆ«csc2xÂ dx = âˆ’cotx+C
- âˆ«secxtanxÂ dx = secx+C âˆ«secxtanxÂ dx = secx+C
- âˆ«cscxcotxÂ dx = âˆ’cscx+C âˆ«cscxcotxÂ dx = âˆ’cscx+C
- âˆ«tanxÂ dx = ln|secx|+C âˆ«tanxÂ dx = ln|secx|+C
- âˆ«cotxÂ dx = ln|sinx|+C âˆ«cotxÂ dx = ln|sinx|+C
- âˆ«secxÂ dx = ln|secx+tanx|+C âˆ«secxÂ dx = ln|secx+tanx|+C
- âˆ«cscxÂ dx = ln|cscxâˆ’cotx|+C âˆ«cscxÂ dx = ln|cscxâˆ’cotx|+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 = 2cos2xâˆ’1 cos2x = 2cos2xâˆ’1 Â so thatÂ cos2x = 12(1+cos2x) cos2x = 12(1+cos2x)
- cos2x = 1âˆ’2sin2x cos2x = 1âˆ’2sin2x Â so thatÂ sin2x = 12(1âˆ’cos2x) sin2x = 12(1âˆ’cos2x)
- cos2x = cos2xâˆ’sin2x cos2x = cos2xâˆ’sin2x
- 1+cot2x = csc2x 1+cot2x = csc2x Â so thatÂ cot2x = csc2xâˆ’1 cot2x = csc2xâˆ’1