Trigonometric Substitution: Method, Formula and Solved Examples

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

Read in Detail: Calculus in Maths

When to Use Trigonometric Substitution?

We use trigonometric substitution in the following cases,

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,

• Integration
• Integration Formulas
• Integration by Substitution
• Integration by Parts

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 sin θ
⇒ 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{0.1 cm}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

Practice Problems on Trigonometric Substitution

1. Evaluate: \int \sqrt{9 - x^2} \, dx

2. Find: \int \frac{dx}{x^2 - 4}

3. Compute: \int \frac{x \, dx}{x^2 + 4}

4. Evaluate: \int \sqrt{x^2 + 16} \, dx

5. Find: \int \frac{dx}{25 - x^2}

6. Compute: \int \sqrt{16 - x^2} \, dx

7. Evaluate: \int \frac{x^2 \, dx}{x^2 + 9}

8. Find: \int \frac{dx}{x^2 + 1}

9. Compute: \int \frac{dx}{x^2 - 1}

10. Evaluate: \int \frac{dx}{4 - x^2}

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Article Tags :

• Mathematics
• School Learning
• Class 12
• Maths-Class-12
• Calculus

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