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Integration By Parts

Integration By Parts

We will be demonstrating a technique of integration that is widely used, called Integration by Part.  

Introduction

Since

$\int e^{x}dx = e^{x} + C$

and

$\int e^{2x}dx = 2e^{2x} + C$

Clearly, 

$\int e^{x}\cdot e^{x}dx \neq \int e^{x}dx\cdot \int e^{x}dx$

In other words,  the integration of the product of two functions is not equal to the product of individual integration of functions.  i.e.

$\int f(x)g(x)dx \neq \int f(x)dx\cdot \int g(x)dx$

Integration by parts is the technique to integrate the functions when typical Integration does not work. For this method, the integrand is of the form

$\int f(x)g(x)dx$

This technique is useful when one function can be differentiated repeatedly, and other function can be integrated repeatedly.  For example, we can apply integration by parts to integrate functions that are products of additional functions, as in finding.

$\int x^2 ln x dx$

Here, the integrand is the product of the functions x2 and ln x . Also, ln x can be differentiated repeatedly, and x2 can be integrated repeatedly. Integration by parts can also apply to the following integrals

$\int x^n\cdot e^{cx}dx$                           $\int x^r\cdot ln ln (cx) dx$   

$\int x^n ( sin\:sin\:cx\:or\:cos\:cos\:cx)dx$                  $\int e^{cx} ⋅(sin\:sin\:cx\: or\: cos\: cos\: cx)dx$

 where c and r are constants and n is a positive integer.

Product Rule In Integral Form

For all differentiable functions f and g. 

$\frac{d}{dx}[f(x)g(x) = f'(x)g(x) + f(x)g'(x)$

                                                                 or        $f(x)g'(x)=\frac{d}{dx}[f(x)g(x)-f'(x)g(x)$

By integrating both sides with respect to x, we have 

$\int f(x)g'(x)dx = \int [\frac{d}{dx}(f(x)g(x))\:-\:f'(x)g(x)]dx$

                                                          =$\int \frac{d}{dx}(f(x)g(x))\:-\:\int f'(x)g(x)dx$

                                                           =$f(x)g(x)\:-\:\int f'(x)g(x)dx$    (By Definition)

                                                     i.e. $\int f(x)g'(x)dx\:=\:f(x)g(x)\:-\:\int f'(x)g(x)dx$

Thus, we have the rule Integration by parts for the integration of functions that are the product of other functions.

In words,

$\int(Integral\: of \: the \: product \: of \: two \: functions)=(first \: function) x (integration \: of \: the \: second \:  function)
– \int [(derivative \: of \: the \: first \: function) × (integration \: of\: the \: second\: function)]dx$ 

Note: Integration by parts formula is only applicable when one function from the product of two functions can be integrated easily.

Steps of This Technique

There are four steps to apply the integration by parts technique. 

  1. Assign functions (f(x),g'(x))
  2. Differentiate and Integrate correct functions
  3. Apply integration by parts formula
  4. Repeat the process if necessary 

Step 1.  Assign functions

The main problem with this formula is to assign functions correctly. There are two possibilities to choose the correct functions for integrating and differentiating. 

i) If one of the functions cannot be integrated, assign first function f(x).

Example i:  $\int$ lnIn x dx

Here, ln x cannot be integrated so, f(x)=ln In x and g'(x)=1

Note: 1 can be considered a function.

ii) If the integration of both functions is possible, assign first function f(x) to the function that eventually differentiates to zero.

Example ii:  $\int x\cdot e^xdx$

Here, both functions can be integrated so we will assign first f(x)=x and g'(x)= ex because x will differentiate to zero.

Step 2.  Differentiate and Integrate correct functions

In step 1, we assigned functions; in step 2, we will differentiate and integrate correct functions. The helpful way is writing functions in a column, the assigned function f(x) will be differentiated, and the other function g'(x) will be integrated.

Example i:  $\int$ In ln x dx

Let       f(x)=ln ln x     and          g'(x)=1

      $\Rightarrow\:\:f'(x)=\frac{d}{dx}In ln x$         $\int g'(x)dx=\int 1dx$

     $\Rightarrow$ f'(x)=1/x                            g(x)=x

Example ii:  $\int$ xexdx

Let       f(x)=x    and           g'(x)=ex

    $\Rightarrow\:\:  f'(x)=\frac{d}{dx}x$         $\int g'(x)dx= \int e^xdx$

     $\Rightarrow$ f'(x)=1                       g(x)=ex

Step 3. Apply integration by parts formula

Substitute assigned functions and differentiated and integrated functions in the formula and simply. 

Example i:  $\int$ In ln x dx

Let       f(x)=ln In x     and         g'(x)=1

    $\Rightarrow\:\:  f'(x)=\frac{d}{dx} In In x$         $\int g'(x)dx= \int 1dx$

      f'(x)=1/x                           g(x)=x

 $\int f(x)g'(x)dx = f(x)g(x) – \int f'(x)g(x)dx$ 

        $\int In ln x 1dx = ln In x× x- \int \frac{1}{x}×xdx$ 

                              =xln In x – $\int$1dx 

                              =xln In x -x+c

Example ii:  $\int$xexdx

Let       f(x)= x    and          g'(x)=ex

     $\Rightarrow\:f'(x)=\frac{d}{dx}x      \int g'(x)dx = \int e^{x}dx$

     $\Rightarrow\:f'(x)=1$                       $g(x) = ex$

$\int f(x)g'(x)dx = f(x)g(x) – \int f'(x)g(x)dx$ 

          $\int xe^{x}dx = xe^{x} – \int 1e^{x}dx$

                            $=xe^x – \int e^xdx$

                            $=xe^x – e^x + c$

Step 4. Repeat the process if necessary

In some integrals may be necessary to repeat the three previous steps several times before reaching a solution.

Example:  $\int$ x2e2xdx

Let      f(x)=x2    and          g'(x)=e2x

$\Rightarrow f'(x) = \frac{d}{dx}x^2$        $\int g'(x)dx = \int e^{2x}dx$

   $\Rightarrow f'(x)=2x$                       $g(x)=\frac{e^{2x}}{2}$

Apply integration by parts formula

$\int f(x)g'(x)dx = f(x)g(x) – \int f'(x)g(x)dx$ 

$\int x^{2}e^{2x}dx = x^{2} \times \frac{e^{2x}}{2} – \int 2x \times \frac{e^{2x}}{2}dx$

                            =$\frac{1}{2}x^{2}e^{2x} – \int xe^{2x}dx$

Repeat the process for $\int$ xe2xdx.

Let       f(x)= x      and        g'(x) = e2x

   $\Rightarrow f'(x) = \frac{d}{dx}x$         $g'(x)dx = \int e^{2x}dx$

    $\Rightarrow f'(x)=1$                      $g(x) = \frac{e^{2x}}{2}$

$\int xe^{2x}dx = x\times \frac{e^{2x}}{2} – \int 1 \times \frac{e^{2x}}{2}dx = \frac{1}{2} xe^{2x} – \frac{1}{4} e^{2x} + c_1$ 

Thus, $\int x^{2}e^{2x}dx = \frac{1}{2} x^{2}e^{2x} – [\frac{1}{2}xe^{2x} – \frac{1}{4}e^{2x} + c_1]$

                                =$\frac{1}{2}x^{2}e^{2x} – \frac{1}{2}xe^{2x} + \frac{1}{4}e^{2x} + c$    where  c = -c1

Rule For the Choice Of First Function

A simple mnemonic to assign the functions for step 1 that we discussed in the previous section. The first function is selected according to the letters in the group LIATE.

LLogarithms Function
IInverse Trigonometric Function
AAlgebraic Function
TTrigonometric Function
EExponential Function

Illustrative Examples

Example (1)  Evaluate $\int xe^{5x}dx‍$

Solution: Observe that

                A:  x is an algebraic function,

                E:  e5x is an exponential function. 

According to the LIATE table, we choose 

               x as first function and e5x as second function.

Let    f(x)=x        and    g'(x)=e5x,

then f'(x)=1       and     g(x)=$\frac{e^{5x}}{5}$   

Apply integration by parts formula, we have 

$\int xe^{5x}dx = x \times \frac{e^{5x}}{5} – \int 1 \times \frac{e^{5x}}{5}dx$

                    =$\frac{1}{5}xe^{5x} – \frac{1}{25}e^{5x}+c$ 

Example (2) Evaluate ‍$\int$ x5 ln In x dx

Solution: : Observe that

                A:  x5 is an algebraic function,

                L:  ln x is a logarithmic function. 

According to the LIATE table, we choose 

                ln x as first function and x5 as second function.

Let     f(x)=lnx     and     g'(x)= x5 ,

then  f'(x)=$\frac{1}{x}$        and       g(x)=$\frac{x^6}{6}$   

Apply integration by parts formula, we have 

$\int$  x5 ln In x dx =‍ $\int$ ln In x x5 dx

                       =(ln In x )$\frac{x^6}{6} – \int \frac{1}{x} \times \frac{x^6}{6}dx$ 

                       =(ln In x )$\frac{x^6}{6} – \frac{1}{6} \int x^5dx$ 

        =$\frac{x^6}{6} In ln x – \frac{1}{6}[\frac{x^6}{6} + c_1]$

        =$\frac{x^6}{6} In ln x – \frac{x^6}{36} + c$      where c = $-\frac{c_1}{6}$

Example (3)  Evaluate ‍$\int$ xcosxdx

Solution: Observe that

                A:  x is an algebraic function,

                T:  cosx is a trigonometric function. 

According to the LIATE table, we choose 

                x as first function and cosx as second function.

Let    f(x)=x    and   g'(x) = cosx,

then  f'(x)=1  and    g(x) = sinx   

Apply integration by parts formula, we have

  ‍xcosxdx = xsinx – $\int$1 X sinxdx

    = xsinx – ( – cosx )+ c

    = xsinx + cosx + c

Choosing u and dv

This section will introduce another notation for the formula of integration by parts that is easy to remember.

Consider u and dv such that 

u = f(x)            and           v = g(x),   

    du = f'(x)dx                      dv = g'(x)dx

Substitute in (1) 

$\int$f(x)g'(x)dx = f(x)g(x) – $\int$ f'(x)g(x)dx

                                                                     $\int$ udv = uv – $\int$ vdu

Now, we will discuss some cases to choose u and dv with example.

Case 1. If function is product of inverse trigonometric function and algebraic function. Then the choice for u and dv will be 

u= algebraic function and dv= inverse trigonometric function.

Example: ‍$\int$ sin-1xdx =‍ $\int$ sin-1x.1dx

We choose u=x     and dv=1dx

                     $\Rightarrow du = \frac{1}{\sqrt{1-x^2}}dx$  and     v=x

Integrating by parts 

$‍x dx = x . x -‍ \int x.\frac{1}{\sqrt{1-x^2}}dx$ 

      =$xx – \int (1-x^2)^{-\frac{1}{2}}(x)dx$ 

  =$xx + \frac{1}{2} \int (1-x^2)^{-\frac{1}{2}}(-2x)dx$

      =$xx + \frac{1}{2} \int (1-x^2)^{-\frac{1}{2}}\frac{d}{dx}(1-x^2)dx$

    =$xx + \frac{1}{2} \frac{(1-x^2)^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} + c$

       =$xx + \frac{1}{2} \frac{(1-x^2)^{\frac{1}{2}}}{1/2} + c$

       =$xx + \sqrt{1-x^2} + c$

Case 2.  If function is product of trigonometric function and algebraic function. Then the choice for u and dv will be

u= algebraic function and dv= trigonometric function.

Example: ‍$\int$ x2sinxdx

We choose u=x2        and  dv = sinxdx

                    $\Rightarrow$ du = 2xdx  and    v = -cosx

Integrating by parts 

            ‍$\int x^2sinxdx = x^2(-cosx)-\int-cosx.2xdx$

                                  =$-x^2cosx + 2\int xcosxdx$

Again we choose u=x        and dv = cosxdx

              $\Rightarrow$     du =1dx   and   v = sinx

Integrating by parts

‍$\int x^2sinxdx = -x^2cosx + 2xsinx -\int sinx\times1dx$

                      =$-x^2cosx + 2xsinx – 2(-cosx) + c$

                      =$-x^2cosx + 2xsinx + 2cosx + c$

Case 3. If function is product of logarithmic function and trigonometric function. Then the choice for u and dv will be

u= logarithmic function and dv= trigonometric function

Example: $\int$ Inln (tan tan x) x dx

We choose u = Inln (tan tan x)         and   dv = x dx

                    $\Rightarrow du = \frac{1}{tantan x}\cdot d x$ and    v = tantan x

Integrating by parts

$\int In ln (tantan x) x dx = lnIn (tantan x)\cdot tantan x – \int tantan x \cdot \frac{1}{tantan x}\cdot x dx$ 

                                       =tantan x Inln (tantan x) – $\int$x dx

                                       =tantan x Inln (tantan x) – tantan x + c

Case 4.  If function is product of algebraic function and logarithmic function. Then the choice for u and dv will be

u= logarithmic function and dv= algebraic function

Example: ‍$\int$x2 Inln x dx

We choose u = Inln x   and dv = x2dx

              $\Rightarrow$du = $\frac{1}{x}$dx and    v = $\frac{x^3}{3}$

Integrating by parts 

‍$\int x^2 lnln x dx =\int lnln\: x x^2dx$

                    =$(lnln x )\frac{x^3}{3} -\int \frac{x^3}{3}\cdot \frac{1}{x}dx$

                      =$\frac{x^3}{3}lnln x -\frac{1}{3}\int x^2dx$

                      =$\frac{x^3}{3}lnln x -\frac{1}{3}[\frac{x^3}{3}+c_1]$

                      =$\frac{x^3}{3}lnln x -\frac{x^3}{9}+c$               where  c = -$\frac{c_1}{3}$

By following the LIATE table, we can choose u and dv easily, and by applying integration by parts formula, we can find the integration of functions that are products of other functions.

Integration By Parts When the Integral of function Reappears

There are some integrals that are unique in that they return on the right-hand side (along with other terms) when the Integration by Parts formula is used. 

Example: $\int$ ecx .{sinsin cx  or coscos cx}dx where c is constant. 

Example  Evaluate ‍‍ $\int$ e-xsinxdx

Solution:  Let I =‍ $\int$e-xsinxdx

Observe that

                A:  ‍e-x is an algebraic function,

                T:  sinx is a trigonometric function. 

According to the LIATE table, we choose 

                sinx as first function and e-x  as second function.

Let     f (x) = sin(x)     and   g'(x) = e-x,

then  f’ (x) = cos(x)  and    g(x) = -e-x   

Apply integration by parts formula, we have

               I = sinx $\times$ (-1)e-x – $\int$cosx $\times$(-1)e-xdx‍

               =-e-xsinx + $\int$e-xcosxdx

Again, repeat the process 

               I= -e-xsinx + cosx $\times$(-1)e-x-$\int$-sinx $\times$(-1)e-xdx‍

               I= -e-xsinx – e-x cosx -‍ $\int$ e-x sinxdx

               I= -e-xsinx – e-x cosx – I + c1   (Integrals repeated)

               I+I= -e-xsinx – e-x cosx + c1

               2I= -e-x(sinx +cosx) + c1

               I=-$\frac{1}{2}$ e-x(sinx +cosx) + c        where  c= $\frac{1}{2}$c1

Example: Let I = $\int$exsinsin x coscos x dx

                            =$\frac{1}{2} \int e^x\cdot$ 2sinsin x coscos x dx 

                            =$\frac{1}{2} \int e^x$ sinsin 2 xdx ∵sinsin 2 x = 2sinsin x coscos x  

We choose u = sinsin 2 x            and   dv = exdx 

              $\Rightarrow$ du = 2coscos 2 xdx   and     v = ex

Integrating by parts

                             I=$\frac{1}{2}[sinsin 2x\cdot e^x – \int e^x \cdot 2coscos 2 xdx]$ 

                             =$\frac{1}{2}e^x sinsin 2x – \int e^x coscos 2 xdx$ 

Again, we choose u = coscos 2 x                  and dv = exdx

              $\Rightarrow$ du = -2coscos 2 xdx   and     v = ex

Again, integrating by parts

                 l=$\frac{1}{2}e^x sinsin 2x – (coscos 2x\cdot e^x – \int e^x \cdot (-2)sinsin 2 xdx )$

                 = $\frac{1}{2}e^x sinsin 2x – (e^x coscos 2x + 2 \int e^x sinsin 2\:xdx)$  (Integral Repeated)

                  = $\frac{1}{2}e^x sinsin 2x – e^x coscos 2x – 4I + c$  

                  $\Rightarrow I + 4I = \frac{1}{2}e^x sinsin 2x – e^x coscos 2x + c$ 

                 $\Rightarrow 5I= \frac{1}{2}e^x sinsin 2x – e^x coscos 2x + c$ 

                  $\Rightarrow I=-\frac{1}{5} e^x coscos 2x + \frac{1}{10} e^x sinsin 2x+\frac{1}{5}c$ 

Definite Integral

Integration by parts can be used with definite integrals.

For all differentiable functions f and g

                  $\int_{a}^{b}f(x)g'(x)dx = [f(x)g(x)]|{a}^{b} -\int{a}^{b}f'(x)g(x)dx$

Example: Evaluate I = $\int_{0}^{\frac{\pi}{2}} x\:sinsin(x)dx$

Solution: 

                  $\int_{0}^{\frac{\pi}{2}} x\:sinsin(x)dx = -x\:coscos x |_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} \:coscos(x)dx$

                  =$0 + sinsin x |_{0}^{\frac{\pi}{2}}$                         
                  =1-0=1

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