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Trigonometric Substitution

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Trigonometric Substitution is one of the many substitution methods of integration where a function or expression in the given integral is substituted with trigonometric functions such as sin, cos, tan, etc. Integration by substitution is a good and easiest approach, anyone can make. It is used when we make a substitution of a function, whose derivative is already included in the given integral function. By this, the function gets simplified, and simple integrals function is obtained which we can integrate easily. It is also known as u-substitution or the reverse chain rule. Or in other words, using this method, we can easily evaluate integrals and antiderivatives. 

Trigonometric Substitution

 

What is Trigonometric Substitution?

Trigonometric substitution is a process in which the substitution of a trigonometric function into another expression takes place. It is used to evaluate integrals or it is a method for finding antiderivatives of functions that contain square roots of quadratic expressions or rational powers of the form \frac{p}{2}           (where p is an integer) of quadratic expressions. Examples of such expressions are

({x^2+4})^\frac{3}{2}             or \sqrt{25-x^2}              or etc.

The method of trigonometric substitution may be called upon when other more common and easier-to-use methods of integration have failed. Trigonometric substitution assumes that you are familiar with standard trigonometric identities, the use of differential notation, integration using u-substitution, and the integration of trigonometric functions.

x = f(θ)

⇒ dx = f'(θ)dθ

Here, we will discuss some important formulae depending on the function we need to integrate, we substitute one of the following trigonometric expressions to simplify the integration:

∫cosx dx = sinx + C

∫sinx dx = −cosx + C

∫sec2x dx = tanx + C

∫cosec2x dx = −cotx + C

∫secx tanx dx = secx + C

∫cosecx cotx dx = −cosecx + C

∫tanx dx = ln|secx| + C

∫cotx dx = ln|sinx| + C

∫secx dx = ln|secx + tanx| + C

∫cosecx dx = ln|cosecx − cotx| + C
 

When to Use Trigonometric Substitution?

We use trigonometric substitution in the following cases,

Expression

Substitution

a2 + x2

x = a tan θ 
OR
x = a cot θ

a2 – x2

x = a sin θ 
OR
x = a cos θ

x2 – a2

x = a sec θ 
OR
x = a cosec θ

\sqrt{\frac{a-x}{a+x}}
OR 
\sqrt{\frac{a+x}{a-x}}

x = a cos 2θ

\sqrt{\frac{x-\alpha}{\beta-x}}
OR
\sqrt{(x-\alpha)(x-\beta)}

x = α cos2θ + β sin2θ

How to Apply Trigonometric Substitution Method?

We can apply the trigonometric substitution method as discussed below,

Integral with a2 – x2

Let’s consider an example of the Integral involving a2 – x2.

Example: \int \frac{1}{\sqrt{a^2-x^2}}\hspace{0.1cm}dx

Lets put, x = a sinθ

⇒ dx = a cosθ dθ

Thus, I = \int \frac{a\hspace{0.1cm}cos\theta \hspace{0.1cm}d\theta}{\sqrt{(a^2-(a\hspace{0.1cm}sin\theta)^2)}}

⇒ I = \int \frac{a\hspace{0.1cm}cos\theta \hspace{0.1cm}d\theta}{\sqrt{(a^2cos^2\theta)}}

⇒ I = \int 1. d\theta

⇒ I = θ + c

As, x = a sinθ

⇒ θ = sin^{-1}(\frac{x}{a})

⇒ I = sin^{-1}(\frac{x}{a}) + c

Integral with x2 + a2

Let’s consider an example of the Integral involving x2 + a2.

Example: Find the integral \bold{\int \frac{1}{x^2+a^2}\hspace{0.1cm}dx}

Solution:

Lets put x = a tanθ

⇒ dx = a sec2θ dθ, we get

Thus, I = \int \frac{1}{(a\hspace{0.1cm}tan\theta)^2+a^2}\hspace{0.1cm}(a\hspace{0.1cm}sec^2\theta \hspace{0.1cm}d\theta)

⇒ I = \int \frac{a\hspace{0.1cm}sec^2\theta \hspace{0.1cm}d\theta}{a^2(sec^2\theta)}

⇒ I = \frac{1}{a}\int 1.d\theta

⇒ I = \frac{1}{a} \theta             + c

As, x = a tanθ

⇒ θ = tan^{-1}(\frac{x}{a})

⇒ I = \frac{1}{a}tan^{-1}(\frac{x}{a})             + c

Integral with a2 + x2.

Let’s consider an example of the Integral involving a2+ x2.

Example: Find the integral of \bold{\int \frac{1}{\sqrt{a^2+x^2}}\hspace{0.1cm}dx}

Solution:

Lets put, x = a tanθ

⇒ dx = a sec2θ dθ

Thus, I = \int \frac{a\hspace{0.1cm}sec^2\theta \hspace{0.1cm}d\theta}{\sqrt{(a^2+(a\hspace{0.1cm}tan\theta)^2)}}

⇒ I = \int \frac{a\hspace{0.1cm}sec^2\theta \hspace{0.1cm}d\theta}{\sqrt{(a^2\hspace{0.1cm}sec^2\theta)}}

⇒ I = \int \frac{a\hspace{0.1cm}sec^2\theta \hspace{0.1cm}d\theta}{a\hspace{0.1cm}sec\theta}

⇒ I = \int sec\hspace{0.1cm}\theta d\theta

⇒ I = log|sec\hspace{0.1cm}\theta+tan\hspace{0.1cm}\theta| + c

⇒ I = log|tan\hspace{0.1cm}\theta+\sqrt{1+tan^2\hspace{0.1cm}\theta}| + c

⇒ I = log|\frac{x}{a}+\sqrt{1+\frac{x^2}{a^2}}|+ c

⇒ I = log|\frac{x}{a}+\sqrt{\frac{a^2+x^2}{a^2}}|+ c

⇒ I = log|\frac{x}{a}+\frac{1}{{a}}\sqrt{a^2+x^2}|+ c

⇒ I = log|x+\sqrt{a^2+x^2}|-log\hspace{0.1cm}a+ c

⇒ I = log|x+\sqrt{a^2+x^2}|+ c_1

Integral with x2 – a2.

Let’s consider an example of the Integral involving x2 – a2.

Example: Find the integral of \bold{\int \frac{1}{\sqrt{x^2-a^2}}\hspace{0.1cm}dx}

Let’s put, x = a secθ

⇒ dx = a secθ tanθ dθ

Thus, I = \int \frac{a\hspace{0.1cm}sec\theta \hspace{0.1cm}tan\theta\hspace{0.1cm}d\theta}{\sqrt{((a\hspace{0.1cm}sec\theta)^2-a^2)}}

⇒ I = \int \frac{a\hspace{0.1cm}sec\theta \hspace{0.1cm}tan\theta\hspace{0.1cm}d\theta}{(a\hspace{0.1cm}tan\theta)}

⇒ I = \int sec\theta\hspace{0.1cm}d\theta

⇒ I = log|sec\hspace{0.1cm}\theta+tan\hspace{0.1cm}\theta| + c

⇒ I = log|sec\hspace{0.1cm}\theta+\sqrt{sec^2\hspace{0.1cm}\theta-1}| + c

⇒ I = log|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}|+ c

⇒ I = log|\frac{x}{a}+\sqrt{\frac{x^2-a^2}{a^2}}|+ c

⇒ I = log|\frac{x}{a}+\frac{1}{{a}}\sqrt{x^2-a^2}|+ c

⇒ I = log|x+\sqrt{x^2-a^2}|-log\hspace{0.1cm}a+ c

⇒ I = log|x+\sqrt{x^2-a^2}|+ c_1

Read More,

Sample Problems on Trigonometric Substitution

Problem 1: Find the integral of \bold{\int \frac{1}{\sqrt{9-25x^2}} \hspace{0.1cm}dx}

Solution:

Taking 5 common in denominator,

⇒ I = \frac{1}{5}\int \frac{1}{\sqrt{\frac{9}{25}-x^2}} \hspace{0.1 cm} dx

⇒ I = \frac{1}{5}\int \frac{1}{\sqrt{(\frac{3}{5})^2-x^2}} \hspace{0.1 cm} dx

According to theorem 1, a = \frac{3}{5}

⇒ I = \frac{1}{5} sin^{-1}(\frac{x}{\frac{3}{5}})              + c

⇒ I = \frac{1}{5} sin^{-1}(\frac{5x}{3})              + c

Problem 2: Find the integral of  \bold{\int \frac{1}{\sqrt{8-2x^2}} \hspace{0.1cm}dx}

Solution:

Taking √2 common in denominator,

⇒ I = \frac{1}{\sqrt{2}}\int \frac{1}{\sqrt{\frac{8}{2}-x^2}} \hspace{0.1 cm} dx

⇒ I = \frac{1}{\sqrt{2}}\int \frac{1}{\sqrt{(2)^2-x^2}} \hspace{0.1 cm} dx

According to theorem 1, a = 2

⇒ I = \frac{1}{\sqrt{2}} sin^{-1}(\frac{x}{2})              +c

⇒ I = \frac{1}{\sqrt{2}} sin^{-1}(\frac{x}{2})              +c

Problem 3: Find the integral of \bold{\int x^3\sqrt{9-x^2}\hspace{0.1cm}dx}

Solution:

By rearranging, we get

\int x^3\sqrt{3^2-x^2}\hspace{0.1cm}dx

Here taking, a = 3 and x = 3 sinθ

⇒ dx = 3 cos θ dθ

Substituting these values,

I = \int (3 sinθ)^3\sqrt{(3^2-(3 sin\theta)^2)}\hspace{0.1cm}3 \hspace{0.1cm}cos \theta\hspace{0.1cm}d\theta

⇒ I = \int 27 sin^3\theta \hspace{0.1cm}3\sqrt{(1-sin^2\theta)}\hspace{0.1cm}3 \hspace{0.1cm}cos \theta\hspace{0.1cm}d\theta

⇒ I = \int 243 \hspace{0.1cm}sin^3\theta cos^2\theta\hspace{0.1cm}d\theta

⇒ I = 243 \int\hspace{0.1cm}sin^2\theta \hspace{0.1cm}sin\theta\hspace{0.1cm}cos^2\theta\hspace{0.1cm}d\theta

⇒ I = 243 \int\hspace{0.1cm}(1-cos^2\theta) \hspace{0.1cm}sin\theta\hspace{0.1cm}cos^2\theta\hspace{0.1cm}d\theta

Lets take,

u = cos θ

⇒ du = -sin θ dθ

Substituting these values, we get

⇒ I = 243 \int\hspace{0.1cm}(1-u^2) \hspace{0.1cm}u^2\hspace{0.1cm}(-du)

⇒ I = -243 \int\hspace{0.1cm}(u^2-u^4) \hspace{0.1cm}du

⇒ I = -243 \int\hspace{0.1cm}u^2 \hspace{0.1cm}du - \int\hspace{0.1cm}u^4 \hspace{0.1cm}du

⇒ I = -243 [\frac{u^3}{3} - \frac{u^5}{5}]

As, u = cos θ and x = 3 sinθ

⇒ cos θ = \sqrt{1-sin^2\theta}

⇒ u = \sqrt{1-(\frac{x}{3})^2}

⇒ u = (1-\frac{x^2}{9})^{\frac{1}{2}}

Hence,I =  -243 [\frac{({(1-\frac{x^2}{9})^{\frac{1}{2}})}^3}{3}-\frac{({(1-\frac{x^2}{9})^{\frac{1}{2}})}^5}{5}]

⇒ I = -243 [\frac{(1-\frac{x^2}{9})^{\frac{3}{2}}}{3}-\frac{(1-\frac{x^2}{9})^{\frac{5}{2}}}{5}]              + c

Problem 4: Find the integral of \bold{\int \frac{1}{4+9x^2} \hspace{0.1cm}dx}

Solution:

Taking 9 common in denominator,

I = \frac{1}{9}\int \frac{1}{\frac{4}{9}+x^2} \hspace{0.1 cm} dx

⇒ I = \frac{1}{9}\int \frac{1}{(\frac{2}{3})^2+x^2} \hspace{0.1 cm} dx

According to theorem 2, a = \frac{2}{3}

⇒ I = \frac{1}{9} \times \frac{1}{\frac{2}{3}}tan^{-1} \frac{x}{(\frac{2}{3})}

⇒ I = \frac{1}{6}tan^{-1} (\frac{3x}{2})+ c

Problem 5: Find the integral of \bold{\int \frac{1}{\sqrt{16x^2+25}}\hspace{0.1cm}dx}

Solution:

Taking 4 common in denominator,

I = \frac{1}{4}\int\frac{1}{\sqrt{x^2+\frac{25}{16}}}

⇒ I = \frac{1}{4}\int\frac{1}{\sqrt{x^2+(\frac{5}{4})^2}}

According to theorem 3, a = \frac{5}{4}

⇒ I = \frac{1}{4}\times log|x+\sqrt{(\frac{5}{4})^2+x^2}|+ c

⇒ I = \frac{1}{4}\times log|\frac{4x+\sqrt{25+16x^2}}{4}|+ c

⇒ I = \frac{1}{4}log|4x+\sqrt{25+16x^2}|-\frac{1}{4}log4+ c

⇒ I = \frac{1}{4}log|4x+\sqrt{25+16x^2}|+ c_1

Problem 6: Find the integral of \bold{\int \frac{1}{\sqrt{4x^2-9}}\hspace{0.1cm}dx}             .

Solution:

Taking 2 common in denominator,

I = \frac{1}{2}\int \frac{1}{\sqrt{x^2-\frac{9}{4}}} \hspace{0.1cm}dx

I = \frac{1}{2}\int \frac{1}{\sqrt{x^2-(\frac{3}{2})^2}} \hspace{0.1cm}dx

According to theorem 4, a = \frac{3}{2}

I = \frac{1}{2}\times log|x+\sqrt{x^2-(\frac{3}{2})^2}|+c

I = \frac{1}{2}log|x+\sqrt{x^2-\frac{9}{4}}|+c

I = \frac{1}{2}log|\frac{2x+\sqrt{x^2-9}}{2}|+c

I = \frac{1}{2}log|2x+\sqrt{x^2-9}|-\frac{1}{2}log2+c

I = \frac{1}{2}log|2x+\sqrt{x^2-9}|+c_1

Problem 7: Find the integral of \bold{\int \frac{1}{x^2-x+1}\hspace{0.1cm}dx}             .

Solution:

After rearranging, we get

I = \int \frac{1}{x^2-x+\frac{1}{4}-\frac{1}{4}+1}\hspace{0.1cm}dx

I = \int \frac{1}{(x-\frac{1}{2})^2+\frac{3}{4})}\hspace{0.1cm}dx

I = \int \frac{1}{(x-\frac{1}{2})^2+(\sqrt{\frac{3}{4}})^2})\hspace{0.1cm}dx

I = \int \frac{1}{(x-\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2})\hspace{0.1cm}dx

According to the theorem 2, we have

x = x-\frac{1}{2}                     and a = \frac{\sqrt{3}}{2}

I = \frac{1}{\frac{\sqrt{3}}{2}} tan^{ -1} \frac{(x-\frac{1}{2})}{\frac{\sqrt{3}}{2}}

I = \frac{2}{\sqrt{3}} tan^{ -1} \frac{(2x-1)}{\sqrt{3}} + c

FAQs on Trigonometric Substitution

Q1: What is Trigonometric Substitution?

Answer:

Trigonometric substitution is technique of integration used to solve the integrals involving  expressions with radicals and square roots such as √(x2 + a2), √(a2 + x2), and √(x2 – a2).

Q2: When should I use Trigonometric Substitution?

Answer:

Trigonometric substitution is useful when you have an integral that involves a radical expression, especially when the radical expression contains a quadratic term.

Q3: What are the Three Trigonometric Substitutions commonly used in Integrals?

Answer:

The three commonly used trigonometric substitutions are:

  • Substitute x = a sin θ when the radical expression contains a term of the form a2 – x2.
  • Substitute x = a tan θ when the radical expression contains a term of the form x2 – a2.
  • Substitute x = a sec θ when the radical expression contains a term of the form x2 + a2.

Q4: How does anyone choose which Trigonometric Substitution to Use?

Answer:

You should choose the trigonometric substitution based on the form of the radical expression. If the radical expression contains a term of the form a^2 – x^2, use x = a sin θ. If the radical expression contains a term of the form x^2 – a^2, use x = a tan θ. If the radical expression contains a term of the form x^2 + a^2, use x = a sec θ.



Last Updated : 07 Jun, 2023
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