Informative line

### 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 $$t.$$

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

#### 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

#### 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

# Problems on Substitution of the type xn = 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 $$t.$$

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

#### 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 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 $$t.$$

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

#### 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

# 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.

#### 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

# 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.

#### 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