Informative line

### Integration By Parts

Learn formula for integration by parts with examples. Use integration by parts formula UV & exponential functions to evaluate indefinite and definite Integrals. Practice problems in which Integration by Parts is to be done more than once.

# Formula for Integration by Parts

• Integration by parts is a technique used for integration of the product of two functions.
• The integration by parts corresponds to the product rule of differentiation just like substitution rule in integration corresponds to chain rule in differentiation.

## Formula for integration by parts

• If $$u(x)$$ and $$v(x)$$ are the two functions of $$x,$$ then by the product rule of differentiation

$$\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$$

$$d(uv)=u\dfrac{dv}{dx}dx+v\dfrac{du}{dx}dx$$

$$d(uv)=udv+vdu$$

Now integrating both sides,

$$\int d(uv)=\int udv+\int vdu$$

$$uv=\int udv+\int vdu$$

$$\int udv=uv-\int vdu$$

This is the formula for integration by parts.

• Another form of formula for integration by parts for two functions $$f(x)$$ and $$g(x)$$  is given as

$$\int f(x)\,g'(x)\,dx=f(x)\,g(x)-\int f'(x)\,g(x)\,dx$$

#### Evaluate $$I=\int x\,sin\,5x\;dx$$ by taking $$f(x) = x$$ and $$g'(x) = sin\; 5x$$.

A $$\dfrac {-x\;cos5x}{5}+\dfrac {sin\;5x}{25}+C$$

B $$\dfrac {x\;cos5x}{5}+C$$

C $$x\, sin\, 5x + x^2 cos\; 5x + C$$

D $$x^3 + x^2 +C$$

×

$$I=\int\;x\;sin\;5x\;dx$$

$$\rightarrow f(x)=x$$, $$g' (x)$$ $$= sin\; 5x$$

We know that

$$\displaystyle \int f(x)\;g'(x)\;dx=f(x)\;g(x)-\int g(x)\;f'(x)\;dx$$

Now,

$$g'(x) = sin \,5x$$

$$\Rightarrow g(x) = \displaystyle\int sin\;5x\;dx = \dfrac {-cos\;5x}{5}$$ (Choose any anti derivative of $$sin\, 5x$$)

$$\therefore$$ $$\displaystyle\int \;x\;sin5x\;dx=\dfrac {-x\;cos\;5x}{5}+\int\dfrac {cos5x}{5}×1\;dx$$

$$=\dfrac {-x\;cos\;5x}{5}+\dfrac {sin\;5x}{25}+C$$

### Evaluate $$I=\int x\,sin\,5x\;dx$$ by taking $$f(x) = x$$ and $$g'(x) = sin\; 5x$$.

A

$$\dfrac {-x\;cos5x}{5}+\dfrac {sin\;5x}{25}+C$$

.

B

$$\dfrac {x\;cos5x}{5}+C$$

C

$$x\, sin\, 5x + x^2 cos\; 5x + C$$

D

$$x^3 + x^2 +C$$

Option A is Correct

# Integration of Product of Two Functions

• The formula of integration by parts is

$$\int udv=uv-\int vdu$$   ... (1)

## How to apply formula

Step 1: First, identify which function is u and which one is dv.

Step 2: Calculate du and v.

Step 3: Put the values in the formula.

Step 4: Evaluate

### Identification of  u and dv

• To identify and dv, we look at the second part of RHS of equation (1).
• We choose u and dv in such order so as to make the integral easy to integrated.

For example: 1. $$\int te^t\,dt$$

Step 1: Identify u and dv.

If we choose et as u and t dt as the dv, then the second part of RHS of equation (1) will be $$\int \dfrac{t^2}{2}e^tdt$$, this is again the product of two functions and it seems complicated too.

Whereas, if we choose t as u and etdt as dv, then the second part of RHS of equation (1) will become $$\int e^t1\cdot dt$$ which is only one function and can be integrated easily.

2. $$\int (\sin x)(\ell n x)\,dx$$

Here, if we choose sin x as u and ln x dx as dv, then the second part of RHS of equation (1) will be the integral of three or more functions and difficult to calculate.

Whereas, if we choose ln x as u and sinx dx as dv, then the second part of RHS of equation (1) will be $$\int \dfrac{1}{x}\cos x\,dx$$ which seems easier than the previous one.

3. $$\int x \sin x\,dx$$

If we choose sinx as u and x dx as dv, then the second part of RHS of equation (1) will be $$\displaystyle\int \cos x\left(\dfrac{x^2}{2}\right)\,dx$$, which will be complicated.

Whereas, if we choose x as u and sinx dx as dv, then the second part of RHS of equation (1) will be $$\int \cos x\,dx$$ which is just a trigonometric function that can be calculated easily.

4. $$\int \cos^{-1}x\,dx$$

Here, we do not know the integral of cos-1x dx. So, we will assume the integrand as the product of 1 and cos-1x dx .

If we choose 1 as u and cos-1x dx as dv, then again we do not know the integral of cos-1x dx and thus, we can't get the value of v.

So, we will choose cos-1x as u and 1.dx as dv.

Conclusion:

• We should choose u as a function which on differentiation reduces the complexity of $$\int vdu$$.
• Also, we should choose dv as a function which on integration reduces the complexity of $$\int vdu$$.
• Let's us solve example (1)

Now, $$I=\int t e^tdt$$

So, t=u and etdt=dv... (2)

Step 2: Find du and v

Step 3: Put the values in the formula.

Example: $$\int \ell nx\,dx$$

Generally we know the direct formula for integration of ln x, but it can't be integrated directly as a single function.

So, we treat this function as a product of 1 and $$\ell nx$$, i.e., $$\int 1\cdot \ell nx \,dx$$

Here, we will derive the formula for integrating the function of ln x using integration by parts formula.

Step 1: Identify u and dv.

If we choose 1 as u and ln x dx as dv, then we don't know the integral of ln x dx. So, we can't get v.

Whereas, if we choose ln x as u and 1.dx as dv, then the second part of RHS of equation (1) will be $$\displaystyle \int x\left(\dfrac{1}{x}\right)\,dx$$ which can be calculated easily.

So here, $$\ell nx=u$$ and $$dv=1\cdot dx$$

Step 2: Find $$du$$ and $$v.$$

Step 3: Put the values in the formula,

This is the required integral of$$\int \ell nx\,dx$$

• Let's take another example.

$$\int x^n\ell nx\,dx$$

Step 1: Identify $$u$$ and $$dv$$.

Here, derivative of $$x^n$$ is $$nx^{n-1}$$

derivative of $$\ell nx$$ is $$\dfrac{1}{x}$$

And anti-derivative of $$x^n$$ is $$\dfrac{x^{n+1}}{n+1}$$

anti-derivative of $$\ell nx$$ is $$x(\ell nx-1)$$

• If we choose $$x^n$$ as $$u$$ and $$\ell nx\,dx$$ as $$dv$$, then the second part of RHS of equation (1) will be $$\int x(\ell nx-1)nx^{n-1}\,dx$$.

This is the product of functions which is very complicated to integrate.

Whereas, if we choose $$x^ndx$$ as $$dv$$ and $$\ell nx$$ as $$u,$$ then the second part of RHS of equation (1) will be $$\displaystyle \int \dfrac{x^{n+1}}{n+1}\,\dfrac{1}{x}\,dx$$ which can be easily calculated.

So, $$u=\ell nx$$ and $$x^ndx=dv$$

Step 2: Find $$du$$ and $$v.$$

Step 3: Put the values in the formula-

Trick to choose  u:

• Many students get confused while choosing $$u$$ and $$v.$$ So, there is a trick to choose $$u.$$

#### Evaluate: $$I=\displaystyle\int t\;e^{2t}\;dt$$

A $$e^t+sin\;t+C$$

B $$\dfrac {e^{2t}}{4}[2t-1]+C$$

C $$2\;cos^2\;t+sin^2\;t+C$$

D $$\dfrac {e^{2t}}{2}[4t-1]+C$$

×

$$I=\displaystyle\int t\;e^{2t}\;dt\rightarrow\;u=t,\;dv=e^{2t}\;dt$$

$$dv=e^{2t}\;dt \Rightarrow\int dv=\int e^{2t}\;dt$$

$$\Rightarrow v=\dfrac {e^{2t}}{2}$$

$$\therefore\;\int\;t\,e^{2t}\;dt=\dfrac{t\;e^{2t}}{2}-\int\,\dfrac{ e^{2t}}{2}×1\;dt$$

$$=\dfrac{t\;e^{2t}}{2}-\dfrac{\;e^{2t}}{4}+C$$

$$=\dfrac{e^{2t}}{4}[2t-1]+C$$

### Evaluate: $$I=\displaystyle\int t\;e^{2t}\;dt$$

A

$$e^t+sin\;t+C$$

.

B

$$\dfrac {e^{2t}}{4}[2t-1]+C$$

C

$$2\;cos^2\;t+sin^2\;t+C$$

D

$$\dfrac {e^{2t}}{2}[4t-1]+C$$

Option B is Correct

# Problems in which Integration by Parts is used more than Once

Integration by parts is a method to integrate the product of two functions.

The formula for integration by parts is:

$$\int\,u\,dv=uv-\int v\, du$$

The second term on R.H.S. of the formulas tells us that $$u$$ is the part that would preferably be differentiated and $$dv$$ is the part that would preferably be integrated.

• In some cases, applying integration by parts once will not solve the problem. We might need to apply it more than once.

Ex. : To integrate the function of the form:

$$\displaystyle I=\int(\text { Polynomial })\;e^{ax}\;dx$$

where $$a\in R$$,

• We know while applying integration by parts method, we choose is u= Polynomial and $$dv=e^{ax}$$ . Because differentiating the polynomial gives simpler term in comparison to if we choose $$dv=(\text { Polynomial })\;dx$$, then its integration will lead to higher power term, where as integration of $$e^{ax}$$ remains the same.

Ex.: If $$u=x^2$$                             and   if  $$dv=x^2\;dx$$

$$\dfrac {du}{dx}=2x$$                                   $$v=\dfrac {x^3}{3}$$

• Thus, $$u=$$ Polynomial and $$dv=e^{x}\;dx$$ is a good choice when we are integrating product of a polynomial and an exponential function.
• On differentiating a polynomial function of degree $$n$$ one time, its power reduces by 1 and we get a polynomial function of degree $$n-1$$

Ex: Let $$u=x^3$$, if we differentiate it once, we get $$du=3x^2\;dx$$

We observe that here power $$3$$ reduces to $$2$$ on differentiating it once.

So, we can conclude that to get the answer, integration by parts method is performed as many times as the degree of polynomial.

Ex. $$\int\;x^2\,e^x\;dx$$

$$u=x^2,\;\;\;\;\;\;\;dv=e^x\;dx$$

$$du=2x\,dx,\;\;\;\;v=e^x$$

On applying integration by parts formula once,

Again we get the product of two functions. So, once again, we apply integration by parts formula.

Let  $$I_1=\int x\,e^x \;dx$$

$$u=x$$                           $$dv=e^x\;dx$$

$$du=dx$$                        $$v=e^x$$

So,

$$I_1=x\,e^x-\int e^x \;dx$$

$$I_1=x\,e^x-e^x$$ .... (ii)

Now, put (ii) in (i)

So,

• In the above example, we saw that the degree of polynomial was 2 (highest power of $$x$$), and we also used the integration by parts, method two times.
• When we have product of two functions will cyclic derivative, then also integration by parts is used more than once.

Ex.: $$\int \,e^x\,cos\,x\;dx$$         or    $$\int \,e^x\,sin\,x\;dx$$

• If we have product of powers of polynomial and logarithmic functions, then also integration by parts is used more than once.

Note: For higher powers of $$x$$ of the form,

$$\int \,x^x\,e^x\,\;dx,\;\int \,x^x\,sin\,x\;dx,\;\int \,x^x\,cos\,x\;dx$$

repeatedly use of integration by parts can evaluate the integrals. Each time, the method reduces the power of $$x$$ by one.

#### Evaluate $$\displaystyle I = \int (x^2+3)\; e^{2x}\;dx$$

A $$e^{2x}[x^2+3x+4]+C$$

B $$\dfrac {e^{2x}}{4}[2x^2-2x+7]+C$$

C $$sin\, x + log \,(tan\, x) + C$$

D $$\dfrac {e^{2x}}{4}[x^2-4x+7]+C$$

×

$$\displaystyle I = \int (x^2+3)\; e^{2x}\;dx$$

$$u = (x^2+3), \;dv = e^{2x}\;dx$$

$$\;dv = e^{2x}\;dx$$ $$\Rightarrow\int dv=\int e^{2x}\;dx$$$$\Rightarrow v=\dfrac {e^{2x}}{2}$$

$$\therefore \;\displaystyle I=\int (x^2+3)\;e^{2x}\,dx=(x^2+3)\dfrac {e^{2x}}{2}-\displaystyle\int\dfrac {e^{2x}}{2}×2x\;dx$$

$$=(x^2+3)\dfrac {e^{2x}}{2}-\underbrace {\int\ {e^{2x}}×x\;dx}_{I_1}$$

$$I_1=\displaystyle\int\ {x\,e^{2x}}\;dx$$

$$u=x\;,dv=e^{2x}\;dx$$

$$\;dv = e^{2x}\;dx$$ $$\Rightarrow\int dv=\int e^{2x}\;dx$$$$\Rightarrow v=\dfrac {e^{2x}}{2}$$

$$\therefore \;I_1=\dfrac {x\,e^{2x}}{2}-\displaystyle\int \dfrac {e^{2x}}{2}×1\;dx =\dfrac {x\,e^{2x}}{2}-\dfrac {e^{2x}}{4}$$

$$\therefore\;I=(x^2+3)\dfrac {e^{2x}}{2}- \left [ \dfrac {x\,e^{2x}}{2}-\dfrac {e^{2x}}{4} \right ]+C$$

$$=(x^2+3)\dfrac {e^{2x}}{2}- \dfrac {x\,e^{2x}}{2}+\dfrac {e^{2x}}{4} +C$$

$$=\dfrac {e^{2x}}{4} [2x^2+6-2x+1]+C$$

$$=\dfrac {e^{2x}}{4} [2x^2-2x+7]+C$$

### Evaluate $$\displaystyle I = \int (x^2+3)\; e^{2x}\;dx$$

A

$$e^{2x}[x^2+3x+4]+C$$

.

B

$$\dfrac {e^{2x}}{4}[2x^2-2x+7]+C$$

C

$$sin\, x + log \,(tan\, x) + C$$

D

$$\dfrac {e^{2x}}{4}[x^2-4x+7]+C$$

Option B is Correct

# Indefinite Integrals Involving Substitution Followed by Integration by Parts

• In some indefinite integrals, first we make an appropriate substitution and then the new integral formed, can be integrated by method of integration by parts.

For e.g.,

Consider

$$I=\int e^{\sqrt x}\;dx$$

If we put $$x = t^2$$, then $$dx = 2t\; dt$$

$$I=\int e^t×2t\;dt$$

$$I=2\int t\,e^t\;dt$$ (This will use integration by parts, so first we use substitution and then integration by parts)

#### First make a substitution and then use integration by parts to evaluate  $$I=\int\,sin\; \sqrt {x}\;dx$$

A $$e^{sin\,x}×\sqrt x+C$$

B $$-2\sqrt {x}\;\;sin\sqrt {x}\;+2cos\;\sqrt {x}\;+C$$

C $$e^{cos\,x}×\sqrt {x}\;+C$$

D $$-2\sqrt {x}\;\;cos\sqrt {x}\;+2\;sin\;\sqrt {x}\;+C$$

×

$$I=\int\;sin\sqrt {x}\;dx$$

Put $$x=t^2$$ so that the integrand gets rationalized

$$\Rightarrow dx=2t\;dt$$

$$\therefore \;I=\int sin\,t×2t\;dt$$

Now use integration by parts

$$u = 2t, \;dv =sint\; dt$$

$$dv =sint\; dt\Rightarrow\int \;dv=\int sin\,t \;dt\Rightarrow v=-cos\,t$$

$$\therefore I=2t×(-cos\,t)-\int -cos\,t ×2\;dt$$

$$=-2t\,cos\,t+2\,sin\,t +C$$

$$=-2\sqrt x\;cos\;\sqrt x+2\;sin\,\sqrt x+C$$ (Put $$t = \sqrt x$$)

### First make a substitution and then use integration by parts to evaluate  $$I=\int\,sin\; \sqrt {x}\;dx$$

A

$$e^{sin\,x}×\sqrt x+C$$

.

B

$$-2\sqrt {x}\;\;sin\sqrt {x}\;+2cos\;\sqrt {x}\;+C$$

C

$$e^{cos\,x}×\sqrt {x}\;+C$$

D

$$-2\sqrt {x}\;\;cos\sqrt {x}\;+2\;sin\;\sqrt {x}\;+C$$

Option D is Correct

#### Choose the option which represents the integral of $$\int \tan^{-1}x\,dx$$.

A $$\tan x+\ell n(1+x^2)+C$$

B $$\tan ^{-1}x+\dfrac{1}{2}\ell n(x^2)+C$$

C $$\tan^{-1}x-\ell n(1+x^2)$$

D $$x \tan^{-1}x-\dfrac{1}{2}\ell n(1+x^2)+C$$

×

Given : $$\int \tan^{-1}x\,dx$$

Here, only one function is given, but we do not know its integral. So, we will use integration by parts formula to find its integral.

We will consider the given integrand as a product of 1 and $$\tan^{-1}x\,dx$$.

$$=\int 1\cdot \tan^{-1}x\,dx$$

First, identifying $$u$$ and $$dv$$.

$$\int 1\cdot \tan^{-1}x\,dx$$

If we choose 1 as $$u$$ and $$\tan^{-1}x\,dx$$ as $$dv$$, then we do not know the integral of $$\tan^{-1}x\,dx$$.

Thus, we can not get $$v.$$

Whereas, if we choose $$\tan^{-1}x$$ as $$u$$ and $$1dx$$ as $$dv$$, then the second part of RHS of integration by parts formula will be $$\int x\dfrac{1}{1+x^2}dx$$ that can be solved by using substitution.

So, $$\tan^{-1}x=u$$ and $$1dx=dv$$

Now, we will find $$v$$ and $$du$$.

$$\tan^{-1}x=u$$

$$\dfrac{d}{dx}(\tan^{-1}x)=\dfrac{du}{dx}$$

$$\dfrac{1}{1+x^2}=\dfrac{du}{dx}$$

$$du=\dfrac{1}{1+x^2}dx$$

$$1\cdot dx=dv$$

Integrating both sides,

$$\int dx=\int dv$$

$$x=v$$

Putting the values in the formula for integration by parts,

$$\int udv=uv-\int vdu$$

$$\displaystyle\int \tan^{-1}x\,dx=x \tan^{-1}x-\int x\dfrac{1}{(1+x^2)}dx$$

Now, let $$\displaystyle I_1=\int x\dfrac{1}{(1+x^2)}\,dx$$

then, $$1+x^2=t$$

$$2x\,dx=dt$$

$$xdx=\dfrac{dt}{2}$$

Putting this in our assumption,

$$\displaystyle I_1=\int\dfrac{1}{t}\left(\dfrac{dt}{2}\right)$$

$$=\displaystyle\dfrac{1}{2}\int \dfrac{1}{t}dt$$

$$=\dfrac{1}{2}\ell n\,|t|=\dfrac{1}{2}\ell n(1+x^2)$$

So, $$\displaystyle\int \tan^{-1}x\,dx=x\tan^{-1} x-\dfrac{1}{2}\ell n(1+x^2)+C$$

Hence, option (D) is correct.

### Choose the option which represents the integral of $$\int \tan^{-1}x\,dx$$.

A

$$\tan x+\ell n(1+x^2)+C$$

.

B

$$\tan ^{-1}x+\dfrac{1}{2}\ell n(x^2)+C$$

C

$$\tan^{-1}x-\ell n(1+x^2)$$

D

$$x \tan^{-1}x-\dfrac{1}{2}\ell n(1+x^2)+C$$

Option D is Correct

# Integration of Product of two Functions with Cyclic Derivative

• Some integrals don't give an answer on applying integration by parts method once.
• In those cases, we apply the method more than once.
• There are some cases, where the original integral on the right hand side of integration by parts formula reappears.
• This type of problem occurs when the product of two functions involve cyclic derivative (since cosine etc.)
• We know that on differentiating sine and cosine functions two times, we again get the same functions, we and cosine respectively.

For Ex.: $$u=sin \,x$$

(1) $$\dfrac {du}{dx}=cos \,x$$

(2) $$\dfrac {d^2u}{dx^2}=-sin \,x$$

• Let us understand by taking another example:

$$I=\int\, e^{ax}\,cos\,bx\;dx$$

• Since the above given integral is a product of two functions, so we use integration by parts method.

Conclusion:

• We should choose $$u$$  as a function which on differentiation reduce the complexity of $$\int v\;du$$
• Also, we should choose $$dv$$ as a function which on integration reduces the complexity of $$\int v\;du$$.

So,

$$u=cos\,bx$$           and      $$dv=e^{ax}\;dx$$

$${u'=-b\,sin\,bx}$$ $$v=\dfrac {e^{ax}}{a}$$

Now, applying integration by parts formula.

$$\int u\,dv=u\,v- \int v\,du$$

$$I=\int e^{ax}\,cos\,bx\,dx=cos\,bx×\dfrac {e^{ax}}{a}-\displaystyle\int\dfrac {e^{ax}}{a}×-b\,sin\,bx\;dx$$

Again we get the product of two functions, so we again, we integration by parts

$$I_1=\int e^{ax}\,sin\,bx\;dx$$

Again, let $$u=sin\,bx$$ ,   $$dv=e^{ax}\;dx$$

$$u'=b\,cos\;bx$$ , $$v=\dfrac {e^{ax}}{a}$$

$$I_1=\dfrac {e^{ax}}{a}\,sin\,bx- \displaystyle\int \dfrac {e^{ax}}{a}× b\,cos\,bx\;dx$$

$$I_1=\dfrac {e^{ax}}{a}\,sin\,bx-\dfrac {b}{a} \displaystyle\int e^{ax}×\,cos\,bx\;dx$$

Here, we can use that $$\displaystyle\int e^{ax} \,cos\,bx\;dx=I$$ appears again,

So,

$$I_1=\dfrac {e^{ax}}{a}\,sin\,bx-\dfrac {b\,I}{a}$$ ... (ii)

Put (ii) in (i),

$$I_1=\dfrac {e^{ax}}{a}\,cos\,bx+\dfrac {b}{a} \left[ \dfrac {e^{ax}}{a} sin\,bx-\dfrac {b}{a} I\right]$$

$$I_1=\dfrac {e^{ax}}{a}\,cos\,bx+\dfrac {b}{a} e^{ax} sin\,bx-\dfrac {b^2}{a^2} I$$

$$I+\dfrac {b^2}{a^2} I= \dfrac{ a\,e^{ax}\,cos\,bx+b\,e^{ax}\,sin\,bx}{a^2}$$

$$\dfrac {a^2I+b^2I}{a^2}= \dfrac{ a\,e^{ax}\,cos\,bx+b\,e^{ax}\,sin\,bx}{a^2}$$

$$I(a^2+b^2)= e^{ax}[a\,cos\,bx+b\,sin\,bx]$$

$$I=\dfrac { e^{ax}[a\,cos\,bx+b\,sin\,bx] } {(a^2+b^2)}$$

$$\Rightarrow \displaystyle\int e^{ax}\,cos\,bx\,dx=\dfrac {e^{ax}}{(a^2+b^2)}[a\, cos\, bx + b\, sin\, bx]$$

Similarly,

$$\Rightarrow \displaystyle\int e^{ax}\,sin\,bx\,dx= \dfrac {e^{ax}}{(a^2+b^2)}[a\, sin\, bx - b\, cos\, bx]$$

#### Evaluate $$I=\int\;e^{2x}\,sin\,x\;dx$$

A $$\dfrac {e^{2x}}{5}(2\,cos\,x-sin\,x)+C$$

B $$e^{x}\,sin\,x+\ell nx+C$$

C $$\dfrac {e^{2x}}{5}(2\,sin\,x-cos\,x)+C$$

D $$\ell n\,(sin\,x)+e^{x}+C$$

×

$$I=\int\;e^{2x}\,sin\,x\;dx$$

$$u=sin\,x\,,\;dv=e^{2x}\;dx$$

$$dv=e^{2x}\;dx$$

$$\Rightarrow \int dv=\int e^{2x}\;dx$$

$$\Rightarrow\; v=\dfrac {e^{2x}}{2}$$

$$\displaystyle\therefore\; I=\int e^{2x}\;sin\,x\;dx=sinx×\dfrac {e^{2x}}{2}-\underbrace {\int \dfrac {e^{2x}}{2}×cos\,x\;dx}_{I_1}$$

Now, $$I_1=\int\;\dfrac {e^{2x}}{2}\,cos\,x\;dx$$

$$u=cos\,x\,,\;dv=\dfrac {e^{2x}}{2}\;dx$$

$$dv=\dfrac {e^{2x}}{2}\;dx$$ $$\Rightarrow\;\displaystyle\int dv=\int\dfrac {e^{2x}}{2}\;dx \Rightarrow v=\dfrac {e^{2x}}{4}$$

$$\therefore$$  $$I_1=cos\,x×\dfrac {e^{2x}}{4}-\displaystyle\int \dfrac {e^{2x}}{4}×(-sin\,x)\;dx$$

$$=(cos\,x)\dfrac {e^{2x}}{4}+\dfrac {1}{4}I$$  (Note that $$I$$ reappears)

$$\therefore$$ $$I=\dfrac {e^{2x}\,sinx}{2}-\left [ \dfrac {e^{2x}\,cosx}{4}+\dfrac {I}{4} \right]$$

$$\Rightarrow\; I+\dfrac {I}{4}=\dfrac {e^{2x}\,sin\,x}{2}- \dfrac {e^{2x}\,cos\,x}{4}$$

$$\Rightarrow\; \dfrac {5I}{4}=\dfrac {e^{2x}}{4}[2sin\,x-cos\,x]$$

$$\Rightarrow\; I=\dfrac {e^{2x}}{5}[2sin\,x-cos\,x]+C$$

### Evaluate $$I=\int\;e^{2x}\,sin\,x\;dx$$

A

$$\dfrac {e^{2x}}{5}(2\,cos\,x-sin\,x)+C$$

.

B

$$e^{x}\,sin\,x+\ell nx+C$$

C

$$\dfrac {e^{2x}}{5}(2\,sin\,x-cos\,x)+C$$

D

$$\ell n\,(sin\,x)+e^{x}+C$$

Option C is Correct

# Definite Integration by Parts

• If we combine the formula for Integration by Parts with Fundamental Theorem of Calculus II, we can find the definite integral by parts.

$$\int f(x)\,g'(x)\;dx=f(x)\;g(x)-\int g(x)\,f'(x)\;dx$$

$$\int\limits_a^b\;f(x)\;g'(x)\;dx=f(x)\,g(x)\Big]_a^b-\int\limits_a^bg(x)\;f'(x)\;dx$$

where both the terms of R.H.S should be finite.

#### Evaluate $$I=\displaystyle\int\limits_1^2\dfrac {\ell n\,x}{x^2}\;dx$$

A $$\dfrac {1}{\ell n\,2}$$

B $$\dfrac {2}{3}\,e$$

C $$\dfrac {1}{2}\,{\ell n} \left (\dfrac {2}{e} \right)$$

D $$\dfrac {1}{2}\,{\ell n} \left (\dfrac {e}{2} \right)$$

×

$$\int\limits_a^b\;f(x)\;g'(x)\;dx=f(x)\,g(x)\Big]_a^b-\int\limits_a^bg(x)\;f'(x)\;dx$$

$$a=1,\; b =2,$$ $$f(x) = \ell n\,x,\; g'(x)=\dfrac {1}{x^2}$$

$$g'(x)=\dfrac {1}{x^2}$$

$$\Rightarrow g(x) = \displaystyle \int \dfrac {1}{x^2}\;dx=\dfrac {x^{-1}}{-1}=\dfrac {-1}{x}$$

$$I=\ell n\;x×\dfrac {(-1)}{x}\Bigg]_1^2-\displaystyle\int\limits_1^2\dfrac {1}{x}×\dfrac {(-1)}{x}\;dx$$

$$=\dfrac {-\ell n\;x}{x}\Bigg]_1^2-\dfrac {1}{x}\Bigg]_1^2=\left ( \dfrac {-\ell n\;2}{2}+0 \right)-\Bigg[\dfrac {1}{2}-1\Bigg]$$

$$=\dfrac {1}{2}-\dfrac {\ell n\;2}{2}= \dfrac {1}{2}(1-\ell n\;2)$$

$$=\dfrac {1}{2}\,\ell n\dfrac {e}{2}$$  (By property of log)

### Evaluate $$I=\displaystyle\int\limits_1^2\dfrac {\ell n\,x}{x^2}\;dx$$

A

$$\dfrac {1}{\ell n\,2}$$

.

B

$$\dfrac {2}{3}\,e$$

C

$$\dfrac {1}{2}\,{\ell n} \left (\dfrac {2}{e} \right)$$

D

$$\dfrac {1}{2}\,{\ell n} \left (\dfrac {e}{2} \right)$$

Option D is Correct

# Integration of Functions involving Powers of Polynomial and Logarithmic Functions

• Integration by parts method is used to integrate the product of two functions.
• To evaluate the integral involving product of the power of polynomial and logarithmic function, we first need to identify u and dv.
• Let us understand by taking an example:

$$I=\int x^m (log\,x)^n\;dx$$

Selection of u and dv:

We know that integration by parts formula is:

$$\int u\,dv=uv-\int v\,du$$

Now if we take u=xand dv=(log x)n dx, then when we integrate dv to find v, the second term becomes very complex.

Whereas, if we take, u=(log x)n and dv=xmdx, then we can simply integrate dv to get v and we also know the derivative of $$log\,x$$, i.e.,

$$\dfrac {d}{dx}\,log\,x=\dfrac {1}{x}$$.

So, $$u=(log\,x)^n$$ and $$dv=x^m\;dx$$ is a good choice.

But to obtain the answer, we have to differentiate $$(log\,x)^n$$ again and again till we get 1 out of it. We know that each time on differentiation the power of $$(log\,x)^n$$ reduces by one.

$$\dfrac {d}{dx}(log\,x)^n=n (log\,x)^{n-1}×\dfrac {1}{x}$$

So, to evaluate this integral, we have to apply integration by parts as many times as the number n.

Ex.:

$$I=\int x^4 (\ell n x)^3\;dx$$

So,

$$u=(\ell n\,x)^3$$  and  $$dv=x^4\;dx$$

$$du=\dfrac {3}{x}(\ell n\,x)^2\;dx$$              $$v=\dfrac {x^5}{5}$$

By integration by parts formula, we have

Again we obtain product of two functions, so by integration by parts formula, we have,

$$I_1=\int x^4 (\ell n x)^2\;dx$$

Let

Now, again on applying integration by parts formula, we have,

$$I_2=\int x^4\,\ell n\,x\;dx$$

Let

Putting (iii) in (ii)

$$I_1=\dfrac {x^5}{5} (\ell nx)^2-\dfrac {2}{5} \left [ \dfrac {x^5}{5}\;\ell n\,x\;-\dfrac {x^5}{25} \right]$$... (iv)

Putting (iv) on (i)

#### Evaluate $$I=\int x^3 (\ell n\,x)^2\;dx$$

A $$\dfrac {x^4}{4}(\ell n\,x)^2-\dfrac {x^4}{8}\ell n\,x+\dfrac {x^4}{32}+C$$

B $$\dfrac {x^4}{8}(\ell n\,x)^2-\dfrac {x^4}{4}\ell n\,x+x^4+C$$

C $$x^3\,e^x+C$$

D $$x\,\ell n\,x+C$$

×

$$I=\int x^3 (\ell n\,x)^2\;dx$$

$$\int f(x)\,g'(x)\;dx=f(x)g(x)-\int f'(x)\,g(x)\;dx$$

$$f(x)=(\ell n\,x)^2, \;g'(x)=x^3$$

$$g'(x)=x^3\Rightarrow g(x)=\int x^3\;dx=\dfrac {x^4}{4}$$

$$\therefore\; I=(\ell n\,x)^2\times \dfrac {x^4}{4}-\int\dfrac {2\ell n\,x}{x}×\dfrac {x^4}{4}\;dx$$

$$= \dfrac {x^4}{4}(\ell n\,x)^2-\dfrac {1}{2}\underbrace{\int \ell n\,x×x^3\;dx}_{I_1}$$

$$I_1=\int x^3\,\ell n\,x\;dx\rightarrow f(x)=\ell n\,x,\;g'(x)=x^3$$

$$g'(x)=x^3\Rightarrow g(x)=\int x^3\;dx=\dfrac {x^4}{4}$$

$$I_1=\ell n\,x× \dfrac {x^4}{4}-\int\dfrac {1}{x}×\dfrac {x^4}{4}\;dx=\dfrac {x^4}{4}\ell n\,x-\dfrac {x^4}{16}$$

$$\therefore\; I=\dfrac {x^4}{4}\;(\ell n\,x)^2 -\dfrac {1}{2} \left [\dfrac {x^4}{4}(\ell n\,x)-\dfrac {x^4}{16}\right]+C$$

$$=\dfrac {x^4}{4}\;(\ell n\,x)^2 -\dfrac {1}{8} x^4(\ell n\,x)+\dfrac {x^4}{32}+C$$

### Evaluate $$I=\int x^3 (\ell n\,x)^2\;dx$$

A

$$\dfrac {x^4}{4}(\ell n\,x)^2-\dfrac {x^4}{8}\ell n\,x+\dfrac {x^4}{32}+C$$

.

B

$$\dfrac {x^4}{8}(\ell n\,x)^2-\dfrac {x^4}{4}\ell n\,x+x^4+C$$

C

$$x^3\,e^x+C$$

D

$$x\,\ell n\,x+C$$

Option A is Correct