Method of substitution: In order to find integrals of functions more effectively, we need techniques that can reduce the given functions to standard forms. Integration by substitution method is one of the essential methods to find the integration. This method is used when complex functions are involved or when direct integration is not possible. The integral of a function can be simplified by using the method of substitution by reducing the given function into a simplified or standard function. This is widely used in topics such as linear algebra and nonlinear systems.

Let us learn more about the integration by substitution method and the steps to solve given problems using this method.

What is Integration by Substitution Method?

Let us recall the concept of integration before understanding the concept of integration by substitution.

The integration of the function \(f\left( x \right)\) is given by \(F\left( x \right)\) and it is represented by

\(\int f (x)dx = F(x) + C\)

Here,

\(F\left( x \right)\) is an anti-derivative

\(f\left( x \right)\) is an integrand

\(dx\) is integrating agent

\(c\) is a constant of integration

\(x\) is a variable of integration

The integration by substitution method is used when the integration of the given function cannot be solved directly. So, the given function can be reduced into standard form by using the proper substitution.

This is also called the ‘Reverse Chain Rule’.

How to Perform the Method of Substitution?

Consider the indefinite integral of a function \(f(x),\,\int f (x)dx\). Here the integral can be reduced to another form by substituting \(x = g(t)\)

Let us see how this works.

Consider, \(I = \int f (x)dx\)

\(x = g(t)\)

Differentiating with respect to \(x\),

\(\frac{{dx}}{{dt}} = {g^\prime }(t)\)

\( \Rightarrow dx = {g^\prime }(t)dt\)

\(\therefore \,I = \int f (x)dx = \int f (g(t)){g^\prime }(t)dt\)

Note:

We should make the substitution for a function whose derivative is also present in the integrand.

Steps to Solve Integration by Substitution Method

The following are the simple five steps that help integrate the method by substitution.

Step 1: Choose a new variable \(t\) for the given function to be reduced.

Step 2: Determine the value of \(dx\) of the given integral, where \(f(x)\) is integrated with respect to \(x\).

Step 3: Make the necessary substitution in the function \(f(x)\), and the new value \(dx\).

Step 4: Integrate the function obtained after substitution.

Step 5: Substitute the value of \(t\) to obtain the final answer.

Standard Results

Let us check some standard results along with proofs, for which we apply the substitution method.

(i) If \(\int f (x)dx = \phi (x)\), then \(\int f (ax + b)dx = \frac{1}{a}\phi (ax + b)\)

Proof: Let \(I = \int f (ax + b)dx\)

Substitute \(ax + b = t\). Then \(d(ax + b) = dt\)

\( \Rightarrow adx = dt\)

\( \Rightarrow dx = \frac{1}{a}dt\)

Therefore, \(I = \int f (ax + b)dx = \frac{1}{a}\int f (t)dt\)

\( = \frac{1}{a}\int f (t)dt\)

\( = \frac{1}{a}\phi (t)\)

\( = \frac{1}{a}\phi (ax + b)\)

(ii) \(\int {{{(ax + b)}^n}} dx = \frac{{{{(ax + b)}^{n + 1}}}}{{a(n + 1)}} + C,\,n \ne – 1\)

Proof: Let \(I = \int {{{(ax + b)}^n}} dx\)

Substitute \(ax + b = t\). Then \(d(ax + b) = dt\)

\( \Rightarrow adx = dt\)

\( \Rightarrow dx = \frac{1}{a}dt\)

Therefore, \(\int {{{(ax + b)}^n}} dx = \frac{1}{a}\int {{t^n}} dt\)

\( = \frac{1}{a} \times \frac{{{t^{n + 1}}}}{{n + 1}} + C\)

\( = \frac{{{{(ax + b)}^{n + 1}}}}{{a(n + 1)}} + C\)

(iii) \(\int {\frac{1}{{ax + b}}} dx = \frac{1}{a}\log |ax + b| + C\)

Proof: Let \(I = \int {\frac{1}{{ax + b}}} dx\)

Substitute \(ax + b = t\). Then \(d(ax + b) = dt\)

\( \Rightarrow adx = dt\)

\( \Rightarrow dx = \frac{1}{a}dt\)

Therefore, \(\int {\frac{1}{{ax + b}}} dx = \frac{1}{a}\int {\frac{1}{t}} dt\)

\( = \frac{1}{a}\log |t| + C\)

\( = \frac{1}{a}\log |ax + b| + C\)

(iv) \(\int {\frac{{{f^\prime }(x)}}{{f(x)}}} dx = \log |f(x)| + C\)

Proof: Let \(I = \int {\frac{{{f^\prime }(x)}}{{f(x)}}} dx\)

Putting \(f(x) = t\). Then \({f^\prime }(x)dx = dt\)

Therefore, \(I = \int {\frac{1}{t}} dt\)

\( = \log \,t + C\)

\( = \log |f(x)| + C\)

Hence, \(\int {\frac{{{f^\prime }(x)}}{{f(x)}}} dx = \log |f(x)| + C\).

Integrals of Trigonometric Functions

(i) \(\int {\tan } \,xdx = – \log |\cos \,x| + C\) or \(\int {\tan } \,xdx = \log |\sec \,x| + C\)

Proof: Let \(I = \int {\tan } \,xdx\). Then, \(I = \int {\frac{{\sin \,x}}{{\cos \,x}}} dx\)

Substitute \(\cos \,x = t\). Then \(d(\cos \,x) = dt\)

\( \Rightarrow – \sin \,xdx = dt\)

\( \Rightarrow dx = – \frac{{dt}}{{\sin \,x}}\)

Therefore,\(I = \int {\frac{{\sin \,x}}{{\cos \,x}}} \times – \frac{{dt}}{{\sin \,x}}\)

\( = – \int {\frac{1}{t}} dt\)

\( = – \log |t| + C\)

\( = – \log |\cos \,x| + C\)

Hence, \(\int {\tan } \,xdx = – \log |\cos \,x| + C\) or \(\int {\tan } \,xdx = \log |\sec \,x| + C\).

(ii) \(\int {\cot } \,xdx = \log |\sin \,x| + C\)

Proof: Let \(I = \int {\cot } \,xdx\). Then \(I = \int {\frac{{\cos \,x}}{{\sin \,x}}} dx\)

Substitute \(\sin \,x = t\). Then \(d(\sin \,x) = dt\)

\( \Rightarrow \cos \,xdx = dt\)

\( \Rightarrow dx = \frac{{dt}}{{\cos \,x}}\)

Therefore, \(I = \int {\frac{{\cos \,x}}{t}} \times \frac{{dt}}{{\cos \,x}}\)

\( = \int {\frac{1}{t}} dt\)

\( = \log |t| + C\)

\( = \log |\sin \,x| + C\)

Hence, \(\int {\cot } \,xdx = \log |\sin \,x| + C\).

(iii) \(\int {\sec } \,xdx = \log |\sec \,x + \tan \,x| + C\)

Proof: Let \(I = \int {\sec } \,xdx\). Then \(I = \int {\frac{{\sec \,x(\sec \,x + \tan \,x)}}{{(\sec \,x + \tan \,x)}}} dx\)

Substitute \(\sec \,x + \tan \,x = t\). Then \(d(\sec \,x + \tan \,x) = dt\)

\( \Rightarrow \left( {\sec \,x\tan \,x + {{\sec }^2}\,x} \right)dx = dt\)

\( \Rightarrow dx = \frac{{dt}}{{\sec \,x(\sec \,x + \tan \,x)}}\)

Therefore, \(I = \int {\frac{{\sec \,x(\sec \,x + \tan \,x)}}{t}} \times \frac{{dt}}{{\sec \,x(\sec \,x + \tan \,x)}}\)

\( = \int {\frac{1}{t}} dt\)

\( = \log |t| + C\)

\( = \log |\sec \,x + \tan \,x| + C\)

Hence, \(\int {\sec \,} xdx = \log |\sec \,x + \tan \,x| + C\).

(iv) \(\int {{\mathop{\rm cosec}\nolimits} } \,xdx = \log |{\mathop{\rm cosec}\nolimits} \,x – \cot \,x| + C\)

Proof: Let \(I = \int {{\mathop{\rm cosec}\nolimits} } \,xdx\). Then \(I = \int {\frac{{{\mathop{\rm cosec}\nolimits} \,x({\mathop{\rm cosec}\nolimits} \,x – \cot \,x)}}{{{\mathop{\rm cosec}\nolimits} \,x – \cot \,x}}} dx\)

Substitute \({\mathop{\rm cosec}\nolimits} \,x – \cot \,x = t\). Then \({\mathop{\rm cosec}\nolimits} \,x – \cot \,x = t\)

\( \Rightarrow \left( { – {\mathop{\rm cosec}\nolimits} \,x\,\cot \,x + {{{\mathop{\rm cosec}\nolimits} }^2}\,x} \right)dx = dt\)

\( \Rightarrow dx = \frac{{dt}}{{{\mathop{\rm cosec}\nolimits} \,x({\mathop{\rm cosec}\nolimits} \,x – \cot \,x)}}\)

Therefore, \(I = \int {\frac{{{\mathop{\rm cosec}\nolimits} \,x({\mathop{\rm cosec}\nolimits} \,x – \cot \,x)}}{{{\mathop{\rm cosec}\nolimits} \,x – \cot \,x}}} \times \frac{{dt}}{{({\mathop{\rm cosec}\nolimits} \,x – \cot \,x){\mathop{\rm cosec}\nolimits} \,x}}\)

\( \Rightarrow I = \int {\frac{1}{t}} dt\)

\( = \log |t| + C\)

\( = \log |{\mathop{\rm cosec}\nolimits} \,x – \cot \,x| + C\)

Hence, \(\int {{\mathop{\rm cosec}\nolimits} \,} xdx = \log |{\mathop{\rm cosec}\nolimits} \,x – \cot \,x| + C\).

Integrals of Some Particular Functions

(i) \(\int {\frac{{dx}}{{{x^2} – {a^2}}}} = \frac{1}{{2a}}\log \left| {\frac{{x – a}}{{x + a}}} \right| + C\)

Proof: We have, \(\frac{1}{{{x^2} – {a^2}}} = \frac{1}{{(x – a)(x + a)}}\)

\( = \frac{1}{{2a}}\left[ {\frac{{(x + a) – (x – a)}}{{(x – a)(x + a)}}} \right]\)

\( = \frac{1}{{2a}}\left[ {\frac{1}{{x – a}} – \frac{1}{{x + a}}} \right]\)

Hence, \(\int {\frac{{dx}}{{{x^2} – {a^2}}}} = \frac{1}{{2a}}\left[ {\int {\frac{{dx}}{{x – a}}} – \int {\frac{{dx}}{{x + a}}} } \right]\)

\( = \frac{1}{{2a}}[\log |(x – a)| – \log |(x + a)|] + C\)

\( = \frac{1}{{2a}}\log \left| {\frac{{x – a}}{{x + a}}} \right| + C\)

(ii) \(\int {\frac{{dx}}{{{a^2} – {x^2}}}} = \frac{1}{{2a}}\log \left| {\frac{{a + x}}{{a – x}}} \right| + C\)

Proof: We have, \(\frac{1}{{{a^2} – {x^2}}} = \frac{1}{{2a}}\left[ {\frac{{(a + x) + (a – x)}}{{(a + x)(a – x)}}} \right]\)

\( = \frac{1}{{2a}}\left[ {\frac{1}{{a – x}} + \frac{1}{{a + x}}} \right]\)

 Hence, \(\int {\frac{{dx}}{{{a^2} – {x^2}}}} = \frac{1}{{2a}}\left[ {\int {\frac{{dx}}{{a – x}}} + \int {\frac{{dx}}{{a + x}}} } \right]\)

\( = \frac{1}{{2a}}[ – \log |a – x| + \log |a + x|] + C\)

 \( = \frac{1}{{2a}}\log \left| {\frac{{a + x}}{{a – x}}} \right| + C\).

(iii) \(\int {\frac{{dx}}{{{x^2} + {a^2}}}} = \frac{1}{a}{\tan ^{ – 1}}\frac{x}{a} + C\)

Proof: Let \(I = \int {\frac{{dx}}{{{x^2} + {a^2}}}} \)

Substitute \(x = a\,\tan \,\theta \). Then \(dx = a\,se{c^2}\,\theta d\theta \)

Therefore, \(\int {\frac{{dx}}{{{x^2} + {a^2}}}} = \int {\frac{{a{{\sec }^2}\,\theta d\theta }}{{{a^2}{{\tan }^2}\,\theta + {a^2}}}} \)

\( = \frac{1}{a}\int d \theta \)

\( = \frac{1}{a}\theta + C\)

\( = \frac{1}{a}{\tan ^{ – 1}}\frac{x}{a} + C\)

(iv) \(\int {\frac{{dx}}{{\sqrt {{x^2} – {a^2}} }}} = \log \left| {x + \sqrt {{x^2} – {a^2}} } \right| + C\)

Proof: Let \(I = \int {\frac{{dx}}{{\sqrt {{x^2} – {a^2}} }}} \)

 Substitute \(x = a\,\sec \,\theta \)

Differentiate with respect to \(x\), then \(dx = a\,\sec \,\theta \,\tan \,\theta d\theta \)

 Therefore, \(\int {\frac{{dx}}{{\sqrt {{x^2} – {a^2}} }}} = \int {\frac{{a\,\sec \,\theta \,\tan \,\theta d\theta }}{{\sqrt {{a^2}\,{{\sec }^2}\,\theta – {a^2}} }}} \)

\( = \int {\sec \,} \theta d\theta \)

\( = \log |\sec \,\theta + \tan \,\theta | + {C_1}\)

\( = \log \left| {\frac{x}{a} + \sqrt {\frac{{{x^2}}}{{{a^2}}} – 1} } \right| + {C_1}\)

\( = \log \left| {x + \sqrt {{x^2} – {a^2}} } \right| – \log |a| + {C_1}\)

 \( = \log \left| {x + \sqrt {{x^2} – {a^2}} } \right| + C\), where \(C = {C_1} – \log |a|\).

(v) \(\int {\frac{{dx}}{{\sqrt {{a^2} – {x^2}} }}} = {\sin ^{ – 1}}\frac{x}{a} + C\)

Proof: Let \(I = \int {\frac{{dx}}{{\sqrt {{a^2} – {x^2}} }}} \)

Substitute \(x = a\,\sin \,\theta \)

Differentiate with respect to \(x\), then \(dx = a\,\cos \,\theta d\theta \)

Therefore, \(\int {\frac{{dx}}{{\sqrt {{a^2} – {x^2}} }}} = \int {\frac{{a\,\cos \,\theta d\theta }}{{\sqrt {{a^2} – {a^2}\,{{\sin }^2}\,\theta } }}} \)

\( = \int d \theta \)

 \( = \theta + C\).

\( = {\sin ^{ – 1}}\frac{x}{a} + C\)

(vi) \(\int {\frac{{dx}}{{\sqrt {{x^2} + {a^2}} }}} = \log \left| {x + \sqrt {{x^2} + {a^2}} } \right| + C\)

Proof: Let \(I = \int {\frac{{dx}}{{\sqrt {{x^2} + {a^2}} }}} \)

 Substitute \(x = a\,\tan \,\theta \)

Differentiate with respect to \(x\), then \(dx = a\,{\sec ^2}\,\theta d\theta \)

Therefore, \(\int {\frac{{dx}}{{\sqrt {{x^2} + {a^2}} }}} = \int {\frac{{a\,{{\sec }^2}\,\theta d\theta }}{{\sqrt {{a^2}\,{{\tan }^2}\,\theta + {a^2}} }}} \)

\( = \int {\sec } \,\theta d\theta \)

\( = \log |(\sec \,\theta + \tan \,\theta )| + {C_1}\)

\( = \log \left| {\frac{x}{a} + \sqrt {\frac{{{x^2}}}{{{a^2}}} + 1} } \right| + {C_1}\)

\( = \log \left| {x + \sqrt {{x^2} + {a^2}} } \right| – \log |a| + {C_1}\)

\( = \log \left| {x + \sqrt {{x^2} + {a^2}} } \right| + C\), where \(C = {C_1} – \log |a|\)

Quick View of Integral Solutions

No.	Integrals	Solutions
1	\(\int {\frac{{{f^\prime }(x)}}{{f(x)}}} dx\)	\(\log |f(x)| + C\)
2	\(\int {\tan } \,xdx\)	\( – \log |\cos \,x| + C\) or \(\int {\tan } \,xdx = \log |\sec \,x| + C\)
3	\(\int {\cot } \,xdx\)	\(\log |\sin \,x| + C\)
4	\(\int {\sec \,} xdx\)	\(\log |\sec \,x + \tan \,x| + C\)
5	\(\int {{\mathop{\rm cosec}\nolimits} } \,xdx\)	\(\log |{\mathop{\rm cosec}\nolimits} \,x – \cot \,x| + C\)
6	\(\int {\frac{{dx}}{{{x^2} – {a^2}}}} \)	\(\frac{1}{{2a}}\log \left| {\frac{{x – a}}{{x + a}}} \right| + C\)
7	\(\int {\frac{{dx}}{{{a^2} – {x^2}}}} \)	\(\frac{1}{{2a}}\log \left| {\frac{{a + x}}{{a – x}}} \right| + C\)
8	\(\int {\frac{{dx}}{{{x^2} + {a^2}}}} \)	\(\frac{1}{a}{\tan ^{ – 1}}\frac{x}{a} + C\)
9	\(\int {\frac{{dx}}{{\sqrt {{x^2} – {a^2}} }}} \)	\(\log \left| {x + \sqrt {{x^2} – {a^2}} } \right| + C\)
10	\(\int {\frac{{dx}}{{\sqrt {{a^2} – {x^2}} }}} \)	\({\sin ^{ – 1}}\frac{x}{a} + C\)
11	\(\int {\frac{{dx}}{{\sqrt {{x^2} + {a^2}} }}} \)	\(\log \left| {x + \sqrt {{x^2} + {a^2}} } \right| + C\)

Important Substitutions in Integration by Substitution method

The following are some important substitutions that help simplify the given function and quickly perform the integration process.

No.	Type of Integrand	Substitutions
1	\(\sqrt {{a^2} – {x^2}} \)	\(x = a\,\sin \,\theta \) or \(x = a\,\cos \,\theta \)
2	\(\frac{1}{{\sqrt {{a^2} – {x^2}} }}\)	\(x = a\,\sin \,\theta \) or \(x = a\,\cos \,\theta \)
3	\(\frac{1}{{\sqrt {{x^2} + {a^2}} }}\)	\(x = a\,tan\,\theta \)
4	\(\frac{1}{{\sqrt {{x^2} – {a^2}} }}\)	\(x = a\,sec\,\theta \)
5	\(\sqrt {{x^2} – {a^2}} \)	\(x = a\,\sec \,\theta \) or \(x = a\,cosec\,\theta \)
6	\({x^2} + {a^2}\) or \(\sqrt {{x^2} + {a^2}} \)	\(x = a\,tan\,\theta \) or \(x = a\,\cot \,\theta \)
7	\(\sqrt {\frac{{a – x}}{{a + x}}} \) or \(\sqrt {\frac{{a + x}}{{a – {x^2}}}} \)	\(x = a\,cos\,2\theta \)

Solved Examples

Q.1. Evaluate \(\int {\frac{{{e^{\tan – 1}}x}}{{1 + {x^2}}}} dx\).
Ans:
Let \(I = \int {\frac{{{e^{{{\tan }^{ – 1}}x}}}}{{1 + {x^2}}}} dx\)
Substitute \({\tan ^{ – 1}}\,x = t\). Then \(\frac{d}{{dx}}\left( {{{\tan }^{ – 1}}\,x} \right) = \frac{d}{{dx}}(t)\)
\( \Rightarrow \frac{1}{{1 + {x^2}}} = \frac{{dt}}{{dx}}\)
\( \Rightarrow dx = \left( {1 + {x^2}} \right)dt\)
Therefore, \(\int {\frac{{{e^{\tan – 1}}x}}{{1 + {x^2}}}} \cdot dx = \int {\frac{{{e^t}}}{{1 + {x^2}}}} \left( {1 + {x^2}} \right)dt\)
\( = \int {{e^t}} dt\)
\( = {e^{{{\tan }^{ – 1}}\,x}} + C\)
Hence, \(\int {\frac{{{e^{{{\tan }^{ – 1}}\,x}}}}{{1 + {x^2}}}} dx = {e^{{{\tan }^{ – 1}}\,x}} + C\).

Q.2. Evaluate \(\frac{1}{{\sqrt {1 + \cos \,2x} }}dx\).
Ans:
Let \(I = \int {\frac{1}{{\sqrt {1 + \cos \,2x} }}} dx\)
\( = \int {\frac{1}{{\sqrt {2\,{{\cos }^2}\,x} }}} dx\)
\( = \frac{1}{{\sqrt 2 }}\int {\sec } \,xdx\)
\( = \frac{1}{{\sqrt 2 }}\log |\sec \,x + \tan \,x| + C\)
Hence, \(\frac{1}{{\sqrt {1 + \cos \,2x} }}dx = \log |\sec \,x + \tan \,x| + C\).

Q.3. Evaluate \(\int {\frac{1}{{\cos (x – a)\cos (x – b)}}} dx\)
Ans:
Let \(I = \int {\frac{1}{{\cos (x – a)\cos (x – b)}}} dx\). Then \(I = \frac{1}{{\sin (a – b)}}\int {\frac{{\sin (a – b)}}{{\cos (x – a)\cos (x – b)}}} dx\)
\( \Rightarrow I = \frac{1}{{\sin (a – b)}}\int {\frac{{\sin ((x – b) – (x – a))}}{{\cos (x – a)\cos (x – b)}}} dx\)
\( \Rightarrow 1 = \frac{1}{{\sin (a – b)}}\int {\frac{{\sin (x – b)\cos (x – a) – \cos (x – b)\sin (x – a)}}{{\cos (x – a)\cos (x – b)}}} dx\)
\( \Rightarrow I = \frac{1}{{\sin \left( {a – b} \right)}}\int {\left\{ {\tan \left( {x – b} \right) – \tan \left( {x – a} \right)} \right\}dx} \)
\( \Rightarrow I = \frac{1}{{\sin (a – b)}}\left| { – {{\log }_e}} \right|\cos (x – b)\left| { + {{\log }_e}} \right|\cos (x – a)|| + C\)
\( \Rightarrow I = \frac{1}{{\sin (a – b)}}{\log _e}\left| {\frac{{\cos (x – a)}}{{\cos (x – b)}}} \right| + C\)
Hence, \(\int {\frac{1}{{\cos (x – a)\cos (x – b)}}} dx = \frac{1}{{\sin (a – b)}}{\log _e}\left| {\frac{{\cos (x – a)}}{{\cos (x – b)}}} \right| + C\).

Q.4. Evaluate \(\int {\frac{{2x + 5}}{{{x^2} + 5x – 7}}} dx\).
Ans:
Let \(I = \int {\frac{{2x + 5}}{{{x^2} + 5x – 7}}} dx\)
Substitute \({x^2} + 5x – 7 = t\). Then \(\frac{d}{{dx}}\left( {{x^2} + 5x – 7} \right) = \frac{d}{{dx}}(t)\)
\( \Rightarrow (2x + 5)dx = dt\)
\( \Rightarrow dx = \frac{{dt}}{{2x + 5}}\)
Therefore, \(I = \int {\frac{{2x + 5}}{{{x^2} + 5x – 7}}} dx = \int {\frac{1}{t}} dt\)
\( = \log |t| + C\)
\( = \log \left| {{x^2} + 5x – 7} \right| + C\)
Hence, \(\int {\frac{{2x + 5}}{{{x^2} + 5x – 7}}} dx = \log \left| {{x^2} + 5x – 7} \right| + C\).

Q.5. Evaluate \(\int {\frac{{{{\sec }^2}\,x}}{{3 + \tan \,x}}} dx\).
Ans:
Let \(I = \int {\frac{{{{\sec }^2}\,x}}{{3 + \tan \,x}}} dx\)
Substitute \(3 + \tan \,x = t\). Then \(\frac{d}{{dx}}(3 + \tan \,x) = \frac{d}{{dx}}(t)\)
\( \Rightarrow {\sec ^2}\,xdx = dt\)
\( \Rightarrow dx = \frac{{dt}}{{{{\sec }^2}\,x}}\)
Therefore, \(I = \int {\frac{{{{\sec }^2}\,x}}{{3 + t}}} \times \frac{{dt}}{{{{\sec }^2}\,x}} = \int {\frac{1}{{3 + t}}} dt\)
\( = \log |3 + t| + C\)
\( = \log |3 + \tan \,x| + C\)
Hence, \(\int {\frac{{{{\sec }^2}\,x}}{{3 + \tan \,x}}} dx = \log |3 + \tan \,x| + C\).

Summary of Method of Substitution

Integrals of certain functions cannot be obtained directly if they are not in one of the standard forms, but they may be reduced to standard forms by proper substitution. This method of evaluating an integral by reducing it to standard form an appropriate substitution is called integration by substitution. If \(\phi (x)\) is a continuously differentiable function, then to evaluate integrals of the form \(\int f (\phi (x)){\phi ^\prime }(x)dx\), we substitute \(\phi (x) = t\) and \({\phi ^\prime }(x)dx = dt\). This substitution reduces the above integral to \(\int f (t)dt\). After evaluating this integral, we substitute the value of \(t\) in the final result. Later explained solved examples with the help of integration by substitution method.

Frequently Asked Questions (FAQs)