Simple Methods to Solve Trigonometric Substitutions

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 = 1−sin^2x sec^2x = 1+tan^2x tan^2x = sec^2x−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,

4−x2√ and (x2+1)3/24−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( θ) ,x = f( θ) ,
dx = f′( θ) d θdx = f′( θ) d θ

example, if
x = sec θ ,x = sec θ ,
Then
dx = sec θtan θ d θdx = sec θtan θ d θ

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
4−x2√4−x2

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Let’s suppose

x = 2sin θ ,x = 2sin θ,
then
4−x2√ = 4−(2sin θ)2√4−x2 = 4−(2sin θ)2
= 4−4sin2 θ√ = 4−4sin2 θ
= 4(1−sin2 θ)√ = 4(1−sin2 θ)
= 4–√⋅1−sin2 θ−−−√ = 4⋅1−sin2 θ

(Recall that  cos2 θ+sin2 θ = 1  cos2 θ+sin2 θ = 1 so that  1−sin2 θ = cos2 θ 1−sin2 θ = cos2 θ.)
= 2⋅cos2 θ√ = 2⋅cos2 θ
= 2⋅∣∣cos θ∣∣ = 2⋅|cos θ|

(Assume that  −π2≤ θ≤π2  −π2≤ θ≤π2 so that  cos θ≥0 cos θ≥0.)
= 2cos θ = 2cos θ
and
dx = 2cos θ d θdx = 2cos θ d θ

Thus,
∫4−x2√dx∫4−x2dx
could be rewritten as
∫4−x2√dx = ∫2cos θ⋅2cos θd θ = 4∫cos2 θd θ∫4−x2dx = ∫2cos θ⋅2cos θd θ = 4∫cos2 θd θ

(Recall that  cos2 θ = 2cos2 θ−1  cos2 θ = 2cos2 θ−1 so that  cos2 θ = 12(1+cos2 θ) cos2 θ = 12(1+cos2 θ) .)
= 4∫12(1+cos2 θ)d θ = 4∫12(1+cos2 θ)d θ
= 2∫(1+cos2 θ)d θ = 2∫(1+cos2 θ)d θ
= 2( θ+12sin2 θ)+C = 2( θ+12sin2 θ)+C
= 2 θ+sin2 θ+C = 2 θ+sin2 θ+C

(Recall that  sin2 θ = 2sin θcos θ sin2 θ = 2sin θcos θ .)
= 2 θ+2sin θcos θ+C = 2 θ+2sin θcos θ+C

Our final response must be expressed in terms of x. Given that x = 2sin x, it follows that
sin θ = x2 = opposite hypotenuse sin θ = x2 = opposite hypotenuse

and
θ = arcsin(x2) θ = arcsin(x2)
Using the given right triangle and the Pythagorean Theorem, we can determine any trig value of θ.

Since
(adjacent)2+(opposite)2 = (hypotenuse)2 ⟶(adjacent)2+(opposite)2 = (hypotenuse)2 ⟶
(adjacent)2+(x)2 = (2)2 ⟶ adjacent = 4−x2√ ⟶(adjacent)2+(x)2 = (2)2 ⟶ adjacent = 4−x2 ⟶

Then
2 θ+2sin θcos θ+C = 2arcsin(x2)+2⋅x2⋅4−x2√22 θ+2sin θcos θ+C = 2arcsin(x2)+2⋅x2⋅4−x22
= 2arcsin(x2)+12x⋅4−x2√+C = 2arcsin(x2)+12x⋅4−x2+C

When use the trig substitution method, we always follow 3 popular trig identities:
(I)  1−sin2 θ = cos2 θ 1−sin2 θ = cos2 θ
(II)  1+tan2 θ = sec2 θ 1+tan2 θ = sec2 θ and
(III)  sec^2 θ−1 = tan2 θ sec2 θ−1 = tan2 θ

For the expression

a2−x2−√a2−x2
we use equation (I) and let
x = asin θx = asin θ

(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

a2−x2−√ = a2−a2sin2 θ−√a2−x2 = a2−a2sin2 θ
= a2(1−sin2 θ)−√ = a2(1−sin2 θ)
= a2cos2 θ−−√ = a2cos2 θ
= a2−−√cos2 θ√ = a2cos2 θ
= acos2 θ√ = acos2 θ
= a∣∣cos θ∣∣ = a|cos θ|
= acos θ = acos θ

and
dx = acos θ d θdx = acos θ d θ

For the expression
a2+x2−√a2+x2
we use equation (II) and let
x = atan θx = atan θ

(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+x2−√ = a2+a2tan2 θ−√a2+x2 = a2+a2tan2 θ
= a2(1+tan2 θ)−√ = a2(1+tan2 θ)
= a2−−√sec2 θ√ = a2sec2 θ
= asec2 θ√ = asec2 θ
= a∣∣sec θ∣∣ = a|sec θ|
= asec θ = asec θ

and
dx = asec2 θ d θdx = asec2 θ d θ

For the expression
x2−a2−√x2−a2

we use equation (3) and let
x = asec θx = asec θ

(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
x2−a2−√ = a2sec2 θ−a2−√x2−a2 = a2sec2 θ−a2
= a2(sec2 θ−1)−√ = a2(sec2 θ−1)
= a2−−√sec2 θ−1−−−√ = a2sec2 θ−1
= atan2 θ√ = atan2 θ
= a∣∣tan θ∣∣ = a|tan θ|
= atan θ = atan θ

and
dx = asec θtan θ d θdx = asec θtan θ d θ

Basic indefinite trigonometric integral formulas :

• ∫cosx dx = sinx+C ∫cosx dx = sinx+C
• ∫sinx dx = −cosx+C ∫sinx dx = −cosx+C
• ∫sec2x dx = tanx+C ∫sec2x dx = tanx+C
• ∫csc2x dx = −cotx+C ∫csc2x dx = −cotx+C
• ∫secxtanx dx = secx+C ∫secxtanx dx = secx+C
• ∫cscxcotx dx = −cscx+C ∫cscxcotx dx = −cscx+C
• ∫tanx dx = ln|secx|+C ∫tanx dx = ln|secx|+C
• ∫cotx dx = ln|sinx|+C ∫cotx dx = ln|sinx|+C
• ∫secx dx = ln|secx+tanx|+C ∫secx dx = ln|secx+tanx|+C
• ∫cscx dx = ln|cscx−cotx|+C ∫cscx dx = ln|cscx−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 = 2cos2x−1 cos2x = 2cos2x−1  so that  cos2x = 12(1+cos2x) cos2x = 12(1+cos2x)
• cos2x = 1−2sin2x cos2x = 1−2sin2x  so that  sin2x = 12(1−cos2x) sin2x = 12(1−cos2x)
• cos2x = cos2x−sin2x cos2x = cos2x−sin2x
• 1+cot2x = csc2x 1+cot2x = csc2x  so that  cot2x = csc2x−1 cot2x = csc2x−1