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Indefinite Integration By Substitution

Learn substitution rule for indefinite integrals with examples. Solving indefinite integration by substitution.

Substitution Rule

  • This rule is used to simplify the given complicated functions like \((\tan x\,\sec^2x),\;\dfrac{\ell nx}{x}\), so that their anti-derivatives can be found easily.
  • This will require some extra observation.
  • These integers are of the form-

\(\displaystyle I=\int f(g(x))\,g'(x)\,dx\)   ... (1)

  • For such integrals, substitution step is necessary.

\(g(x)=t\)   [substitution step]

Differentiate both sides with respect to \(x\)

\(\dfrac{d}{dx}(g(x))=\dfrac{dt}{dx}\)

\(g'(x)=\dfrac{dt}{dx}\)

\(g'(x)\,dx=dt\)

Substitute this value in equation (1)

\(\displaystyle I=\int f(\underbrace{g(x)}_{t})\,\underbrace{g'(x)\,dx}_{dt}\)

\(\displaystyle I=\int f(t)\,dt\)

  • If anti-derivative of \(f\) is known, we can find the above integral and then again substitute \(t=g(x)\), to get the answer in terms of \(x\).

Examples:

\(1.\;\displaystyle I=\int 3x^2(\sin x^3)\,dx\)

Step 1: Choose the proper substitute.

Here, it can be observed that \(3x^2\) is the derivative of \(x^3\).

So, we assume \(x^3=t\)

Step 2: Differentiate both sides with respect to \(x\) and find \(dt\).

\(\dfrac{d}{dx}(x^3)=\dfrac{dt}{dx}\)

\(3x^2=\dfrac{dt}{dx}\)

\(3x^2\,dx=dt\)

Step 3: Substitute these values in the given integral.

\(\displaystyle I=\int\sin t\,dt\)

Step 4: \(\displaystyle I=\int\sin x\,dx=-\cos x+C\)

So, \(I=-\cos t+C\)

Step 5: Substitute the values of \(t\)

\(I=-\cos x^3+C\)

\(2.\,\;\displaystyle I=\int\dfrac{4x}{2x^2+3}dx\)

Step 1: Choose the proper substitute.

Here, it can be observed that \(4x\) is the derivative of \(2x^2+3\).

So, \(2x^2+3=t\)

Step 2: Differentiate both sides with respect to \(x\) and find \(dt\).

\(\dfrac{d}{dx}(2x^2+3)=\dfrac{dt}{dx}\)

\(4x=\dfrac{dt}{dx}\)

\(4x\,dx=dt\)

Step 3: Substitute these values in the given integral.

\(\displaystyle I=\int\dfrac{1}{t}\,dt\)

Step 4: \(\displaystyle I=\int\dfrac{1}{x}\,dx=\ell n|x|+C\)

So, \(I=\ell n|t|+C\)

Step 5: Substitute the value of \(t.\)

\(I=\ell n|2x^2+3|+C\)

Illustration Questions

Find  \(I=\int x^2\;sin(x^3)\;dx\)

A \(\dfrac {2}{3}\;sin^3x+C\)

B \(\dfrac {1}{3}\;tanx+C\)

C \(\dfrac {2}{3}\;cos^2x+C\)

D \(\dfrac {-1}{3}\;cos\;x^3+C\)

×

\(I=\int x^2\;sin(x^3)\;dx\)

Observe a function and its derivative \(\dfrac {d}{dx}(x^3)=3x^2\).

 

\(I=\dfrac {1}{3}\int \underbrace{3x^2\;sin(x^3)\;dx}_\text {Constant 3 can always be multiplied.}\)

Put \(x^3=t \)

\(\Rightarrow3x^2\;dx=dt\)

\(\therefore \;I=\dfrac {1}{3}\int sin\;t\;dt\)

\( =\dfrac {-1}{3}\;cos\,t+C\)

\(=\dfrac {-1}{3}cos\,x^3+C\)  (put back \(t = x^3\))

Find  \(I=\int x^2\;sin(x^3)\;dx\)

A

\(\dfrac {2}{3}\;sin^3x+C\)

.

B

\(\dfrac {1}{3}\;tanx+C\)

C

\(\dfrac {2}{3}\;cos^2x+C\)

D

\(\dfrac {-1}{3}\;cos\;x^3+C\)

Option D is Correct

Illustration Questions

Identify the correct substitute for the integral \(I=\displaystyle\int\dfrac {sin\sqrt x}{\sqrt x}\;dx\)  

A \(\sqrt {x}=t\)

B \(sinx=t\)

C \(x^2=t\)

D \(sin^2x=t\)

×

Observe a function and its derivative which are both present in the integrand.

 \(\sqrt x=t\\ \Rightarrow\;\dfrac {1}{2\sqrt x}dx=dt\)

\(\Rightarrow\;\dfrac {1}{\sqrt x}dx=2dt\)

\(\therefore \;\) Option (A) is correct.

Identify the correct substitute for the integral \(I=\displaystyle\int\dfrac {sin\sqrt x}{\sqrt x}\;dx\)  

A

\(\sqrt {x}=t\)

.

B

\(sinx=t\)

C

\(x^2=t\)

D

\(sin^2x=t\)

Option A is Correct

Trigonometric Indefinite Integrals using Substitution

  • There are integrals which have \(sine\) and \(cosine\) functions with powers.
  • Generally, the form is \(\int(\sin\theta)^m\,(\cos\theta)^n\,d\theta\),

where \(m\;\&\;n\) are whole numbers.

  • For such type of integrals, there are specific substitutions depending upon the values of \(m\) and \(n\) , where \(m\) and \(n\) are whole numbers.

1. When \(m\) is odd,

substitute \(\cos \theta=t\)

2. When \(n\) is add,

substitute \(\sin \theta=t\)

3. When both \(m\) and \(n\) are odd,

substitute \(\sin \theta=t\)

Reason for such substitutions

  • These substitution are based on the fact that even powers of \(sine\) and \(cosine\) functions can be expressed in terms of the other without any radical sign.

\(\sin^2\theta=-1\cos^2\theta\;\text{and}\;\cos^2\theta=1-\sin^2\theta\)

  • This will become more clear through following examples.

Examples: \(I=\int\sin^3\theta\;\cos^2\theta\;d\theta\)

Step 1: Here, power of \(\sin\theta\), i.e., \(m\) is odd.

So, we will substitute \(\cos\theta=t\)

Step 2: Differentiate both sides with respect to \(\theta\)

\(-\sin\theta=\dfrac{dt}{d\theta}\)

\(-\sin\,d\theta=dt\)

Step 3: Substitute the values in the integral.

Here, \(\sin^3\theta\) can be written as \(1-\cos^2\theta\)

So, \(\displaystyle I=\int(1-\underbrace{\cos^2\theta}_{t^2})\;\underbrace{\cos^2\theta}_{t^2}\;\underbrace{\sin\theta\;d\theta}_{-dt}\)

\(I=\int(1-t^2)t^2\,(-dt)\)

Step 4: Integrate the integral

\(I=-\int(t^2-t^4)\,dt\)

\(I=-\int t^2dt+\int t^4dt\)

\(I=\dfrac{-t^3}{3}+\dfrac{t^5}{5}+C\)

Step 5: Substitute the value of \(t\) in the result.

\(I=\dfrac{-\cos^3\theta}{3}+\dfrac{\cos^5\theta}{5}+C\)

This is the required integral.

Illustration Questions

Find  \(I=\int sin^6\theta\; cos\theta\;d\theta\)

A \(\dfrac {(cos\,\theta)^6}{5}+C\)

B \(tan^3\,\theta+C\)

C \(\dfrac {(sin\,\theta)^7}{7}+C\)

D \(\dfrac {(sin\,\theta)^4}{5}+C\)

×

\(I=\int sin^6\theta\; cos\theta\;d\theta\)

Observe a function \(sin\;\theta\) and its derivative \(\dfrac {d}{d\theta}\;sin\theta=cos\theta\) both are present in the integrand function.

Put \(sin\;\theta=t\) 

\(\Rightarrow \;cos\theta=\dfrac {dt}{d\theta}\)

\(\Rightarrow \;cos\theta\;{d\theta}={dt}\)

\(\displaystyle I=\int t^6\;dt=\dfrac {t^7}{7}+C\)

\(=\dfrac {(sin\;\theta)^7}{7}+C\)  (put back \(t=sin\;\theta\))

Find  \(I=\int sin^6\theta\; cos\theta\;d\theta\)

A

\(\dfrac {(cos\,\theta)^6}{5}+C\)

.

B

\(tan^3\,\theta+C\)

C

\(\dfrac {(sin\,\theta)^7}{7}+C\)

D

\(\dfrac {(sin\,\theta)^4}{5}+C\)

Option C is Correct

Problems on Substitution of the type x= t (where n is not a Positive Integer)

  • In some integrand the expression \(x^n\) and its derivative \(n\,x^{n-1}\) both are present but when \(n\) is not an integer, they are not easy to identify.

For this, we must remember that if \(\sqrt x\) and its derivative \(\dfrac{1}{2\sqrt x}\)

\(\Big(\) i.e., \(x^{\frac{1}{2}}\;\text{and}\;x^{\frac{-1}{2}}\Big)\), are present in the integral, only then we can substitute \(\sqrt x=t\).

For example:

\(\displaystyle1.\;I=\int\dfrac{e^{\sqrt x}}{\sqrt x}dx\)

Here, the function \(\sqrt x\) and its derivative \(\dfrac{1}{2\sqrt x}\), both are present.

So, substitute \(\sqrt x=t\)

\(\dfrac{1}{2\sqrt x}=\dfrac{dt}{dx}\)

\(\dfrac{1}{2\sqrt x}dx=dt\)

\(\dfrac{dx}{\sqrt x}=2dt\)

So, \(\displaystyle I=\int\dfrac{e^{\sqrt x}dx}{\underbrace{\sqrt x}_{2dt}}\)

\(I=\int e^t\,2dt\)

\(I=2\int e^t\,dt\)

\(I=2e^t+C\)

Now, substitute the value of \(t.\)

\(I=2e^{\sqrt x}+C\)

\(\displaystyle 2.\;I=\int\sqrt x\,cosec^2(x\sqrt x)\,dx\)

Here, the function \(x\sqrt x\) and its derivative \(\dfrac{3\sqrt x}{2}\), both are present.

So, substitute \(x\sqrt x=t\)

\(\dfrac{3\sqrt x}{2}=\dfrac{dt}{dx}\)

\(\sqrt x\,dx=\dfrac{2}{3}dt\)

So, \(\displaystyle I=\int cosec^2 \,\underbrace{\sqrt[x]x}_{t}\;\underbrace{\sqrt x\,dx}_{\frac{2}{3}dt}\)

\(\displaystyle I=\int cosec^2\,t\dfrac{2}{3}dt\)

\(I\displaystyle =\dfrac{2}{3}\int cosec^2\,t\,dt\)

\(I=\dfrac{2}{3}(-\cot t)+C\)

\(I=I=\dfrac{-2}{3}\cot t+C\)

Now, substitute the value of \(t.\)

\(I=\dfrac{-2}{3}-\cot (x\sqrt x)+C\)

Illustration Questions

Find  \(I=\displaystyle \int\dfrac {sec^2\,\sqrt x}{\sqrt x}\;dx\)

A   \(2\,sinx\,\sqrt x+C\)

B   \(cos\dfrac {2}{x}+C\)

C   \(2\,tan\,\sqrt x+C\)

D   \(sin^2\dfrac {x}{2}+C\)

×

\(I=\displaystyle\int\,\dfrac {sec^2\,\sqrt x}{\sqrt x}\;dx\)

Observe that function \(\sqrt x\) and  its derivative  \(\dfrac {d}{dx}\sqrt x=\dfrac {1}{2\sqrt x}\)  both are present in the integrand function.

\(I=2\displaystyle\int\dfrac {1}{2\sqrt x}\,sec^2\,\sqrt x\;dx\)

Put \(\sqrt x = t\) 

\(\Rightarrow\;\dfrac {1}{2\sqrt x}dx=dt\)

\(I=2\int sec^2\;t\;dt\\=2\;tan\;t+C\)

  \(=2\;tan\sqrt x+C\)  (by putting \(t=\sqrt x\))

Find  \(I=\displaystyle \int\dfrac {sec^2\,\sqrt x}{\sqrt x}\;dx\)

A

  \(2\,sinx\,\sqrt x+C\)

.

B

  \(cos\dfrac {2}{x}+C\)

C

  \(2\,tan\,\sqrt x+C\)

D

  \(sin^2\dfrac {x}{2}+C\)

Option C is Correct

Integrals of the Function

  • \(\int f(ax+b)\;dx\), where \(\int f(x)\;dx\) is known
  • To evaluate \(\int f(ax+b)\;dx\) put \(ax + b = t\)

\(\Rightarrow\;ax\;dx=dt\)

\(\Rightarrow\;dx=\dfrac {dt}{a}\)

\(I\) becomes\(\rightarrow\) \(I=\displaystyle\int\;f(t) \dfrac {dt}{a}=\dfrac {1}{a}\int f(t)\;dt\)

Now integrate and back substitute \(t=ax+b\)

For example:

Let the function be

\(\displaystyle I=\int\cos\,(4x+7)\,dx\)

Step 1: Substitution step

\(4x+7=t\)

Step 2: Differentiate with respect to \(x\)

\(4=\dfrac{dt}{dx}\)

\(4dx=dt\)

Here, only \(dx\) appears in\(I\),

So, we will put

\(dx=\dfrac{dt}{4}\)

Step 3: Substitute the values in the integral,

\(\displaystyle I=\int\cos t\,\dfrac{dt}{4}\)

Step 4: Simplify integrate it.

\(\displaystyle I=\dfrac{1}{4}\int\cos t\,dt\)

\(I+\dfrac{1}{4}\sin t+C\)

Step 5: Substitute the value of \(t.\)

\(I=\dfrac{1}{4}\sin(4x+7)+C\)

This is the required integral.

Illustration Questions

Find  \(I=\displaystyle\int(4-3x)^7\;dx\)

A \(\dfrac {-(4-3x)^6}{7}+C\)

B \(\dfrac {(3+2x)^8}{8}+C\)

C \(\dfrac {-(4-3x)^8}{24}+C\)

D \(\dfrac {(4+7x)^7}{8}+C\)

×

\(I=\displaystyle\int(4-3x)^7\;dx\)

put \(4-3x=t\)

\(\Rightarrow-3dx=dt\)

\(\Rightarrow\; dx=\dfrac {-dt}{3}\)

\(I=\displaystyle\int t^7×-\dfrac {1}{3}\;dt\)

\(=-\dfrac {1}{3}\displaystyle\int t^7\;dt\)

\(=-\dfrac {1}{3}\;\dfrac {t^8}{8}+C\\=\dfrac {-t^8}{24}+C\)

By putting \(t=4-3x\)

\(=\dfrac {-(4-3x)^8}{24}+C\)

Find  \(I=\displaystyle\int(4-3x)^7\;dx\)

A

\(\dfrac {-(4-3x)^6}{7}+C\)

.

B

\(\dfrac {(3+2x)^8}{8}+C\)

C

\(\dfrac {-(4-3x)^8}{24}+C\)

D

\(\dfrac {(4+7x)^7}{8}+C\)

Option C is Correct

Integrals in which the Substitution is non-obvious

  • The integrals in which \(x^n\) and its derivative \(n\,x^{n-1}\), both are present, can be evaluated by putting \(x^n=t\).
  • Similarly, if the function of the form, \(a+bx^n\) and its derivative \(x^{n-1}\), both are present, we can make the substitution as-

\(x+bx^n=t\)

Here, \(a\) and \(b\) are constants.

For example:

\(1.\;I=\displaystyle\int\dfrac {x^3}{(1+2x^4)^3}\;dx\)

put \(1+2x^4=t\)  

\(\Rightarrow\;8x^3\;dx=dt\)

\(\therefore\) \(I=\displaystyle\dfrac {1}{8}\int\dfrac {dt}{t^3}\\=\dfrac {1}{8}\int\;t^{-3}dt\)

\(=\dfrac {1}{8}×\dfrac {t^{-2}}{-2} + C\\=\dfrac {-1}{16t^2}+C\)

\(=\dfrac {-1}{16(1+2x^4)^2}+C\)

\(2.\;I=\displaystyle\int\dfrac {(x+1)}{\sqrt{x^2+2x+3}}\;dx\)

Here, substitution step will be

\(x^2+2x+3=t\)

Because the function \(x^2+2x+3\) and its derivative \(2(x+1)\), both are present.

Derivative both sides with respect to \(x\)

\(2x+2=\dfrac{dt}{dx}\)

\(2(x+1)dx=dt\)

\((x+1)dx=\dfrac{dt}{2}\)

Substitute these values in the integral,

\(I=\displaystyle\int\left(\dfrac {1}{\underbrace{\sqrt{x^2+2x+3}}_{t}}\right)\;\underbrace{(x+1)dx}_{\frac{dt}{2}}\)

\(I=\displaystyle\int\dfrac {1}{t^{\frac{1}{2}}}\;\dfrac{dt}{2}\)

\(=\dfrac{1}{2}\displaystyle\int t^{\frac{-1}{2}}dt\)

\(=\dfrac{1}{2}\;\dfrac{t^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+C\)

\(=\dfrac{2}{2}\;t^{\frac{1}{2}}+C\)

\(=\sqrt t+C\)

Substitute the value of \(t.\)

\(I=\sqrt{x^2+2x+3}+C\)

Illustration Questions

Find  \(I=\int\;(4x+2)\sqrt {x^2+x+1}\;dx\)

A \(\dfrac {4}{3}(x^2+x+1)^{3/2}+C\)

B \(\dfrac {5}{2}(x^2+x+1)^{5}+C\)

C \(\dfrac {3}{4}(x^2+x+1)^{5/2}+C\)

D \(\dfrac {2}{3}\;sin^2x+C\)

×

\(I=\int\;(4x+2)\sqrt {x^2+x+1}\;dx\)

Observe that function \((x^2+x+1)\)  and its derivative  \(\dfrac {d}{dx}(x^2+x+1)=2x+1\) both are present in the integrand.

 

\(I=2\int\;(2x+1)\sqrt {x^2+x+1}\;dx\)

put \(x^2+x+1=t\) 

\(\Rightarrow \;(2x+1) dx = dt\)

\(I = 2 \int\; \sqrt {t}\;dt\\=2\int \;t^{1/2}\;dt\)

\(=\dfrac {2\,t^{3/2}}{3/2}+C\)

\(=\dfrac {4}{3}\;t^{3/2}+C\)

\(=\dfrac {4}{3}(x^2+x+1)^{3/2}+C\)        (by putting \(t=(x^2+x+1)\) )

Find  \(I=\int\;(4x+2)\sqrt {x^2+x+1}\;dx\)

A

\(\dfrac {4}{3}(x^2+x+1)^{3/2}+C\)

.

B

\(\dfrac {5}{2}(x^2+x+1)^{5}+C\)

C

\(\dfrac {3}{4}(x^2+x+1)^{5/2}+C\)

D

\(\dfrac {2}{3}\;sin^2x+C\)

Option A is Correct

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