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Trigonometric Substitution

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  • Difficulty Level : Medium
  • Last Updated : 07 Apr, 2021
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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. 

The integration by substitution method:

Let us take an integral I = ∫f(x)dx  …..(i)

Now we can transform this integral by changing its independent variable x to u by substituting x = g(u).

Differentiate g(u) w.r.t.x, we get

dx/du = g'(u)

or we can write as

dx = g'(u) du

Now put all these values in eq(i), we get

∫f(x)dx = ∫f(g(u)).g'(u) du

Hence, the general form of integration by substitution is 

∫f(g(x)).g′(x).dx = ∫f(u).dt

Here u = g(x). The integration by substitution are of two types:

1. Direct Substitution: The direct substitution follows the integration by substitution theorem. 

Theorem:

Let us assume f be the function defined in I interval and has an inverse function F. Let g be the function from H interval into I interval which is differentiable on H. So, F(g) is an inverse function or antiderivative of f(g)g’ on H:

∫f(g(x)).g′(x).dx = ∫f(y).dy\y = g(x)

In direct substitution, if the given integral is in the form of ∫f(g(x)).g′(x).dx(by rearranging). Then we can transform this integral by changing its independent variable x to u by substituting u = g(x).

∫f(g(x)).g′(x).dx = ∫f(u).g′(x).dx

Now we have to replace g′(x)dx with du.

So we get

∫f(g(x)).g′(x).dx = ∫f(u).du

Now the final result is in this form

F(u) + C = F(g(x)) + C 

Example: sin(x2)8xdx

As, here it is 8x not 2x. 

Then we will try to make it fit into the definition. We create its derivative here and do some rearranging of functions.

∫4sin(x2)2xdx

= 4 \int sin(u)\hspace{0.1cm}du

= 4 (- cos(u) + c)

= – 4 cos(u) + c1  (c1 = 4c)

Now, substituting u = x2, we get

sin(x2)8xdx = – 4 cos(x2) + c1

2. Indirect Substitution: It is also a method which includes substitution for simplification, but it reverses of direct substitution. 

In indirect substitution, if the given integral is in the form of ∫f(u).du. Then we can transform this integral back to its original independent variable u to x by substituting u = g(x).

 ∫f(u).du = f(g(x))du 

Now we have to replace du with g′(x)dx.

So we get

f(g(x))du = f(g(x))g′(x)dx 

Now the final result is in this form

G(u) + C = G(g-1(x)) + C 

As, It has been mentioned earlier that it looks as if we followed the above direct substitution in the opposite direction. Indirect substitution is used quite rarely, but there are specific types of integrals. Mostly it is useful where integrals with roots take place.

Example: \int x^2 \sqrt{x^3+5} \hspace{0.1cm}dx

Let’s substitute

u = \sqrt{x^3+5}

u2 = x3 + 5

Differentiating, w.r.t to x, we get

2u du/dx = 3x2

x2 dx = 2/3u du

By substituting these values, we get

\int u \hspace{0.1cm}(\frac{2}{3}u du)

\frac{2}{3} \int u^2 \hspace{0.1cm}du

\frac{2}{3} (\frac{u^3}{3} + c)

2/9 u3+ c1 (c1 = 2/9 c)

As, u = \sqrt{x^3+5} , we get 

I = \frac{2}{9} (\sqrt{x^3+5})^3  + c1

Trigonometric substitution

Trigonometric substitution is a process in which substitution 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
 

Following are some substitutions useful in evaluating integrals:

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 = ∝ cos2θ + β sin2θ

Integrands consisting a2 – x2.

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

Lets put, x = a sinθ

dx = a cosθ dθ

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

Integrands consisting x2 + a2.

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

Lets put x = a tanθ

dx = a sec2θ dθ, we get

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

Integrands consisting a2 + x2.

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

Lets put, x = a tanθ

dx = a sec2θ dθ

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

Integrands consisting x2 – a2.

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

Let’s put, x = a secθ

dx = a secθ tanθ dθ

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

Sample Problems

Question 1. \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

Question 2.  \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

Question 3. \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

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

Question 4. \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

Question 5. \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

Question 6. \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

Question 7. \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


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