Integration by Parts

Integration by Parts is a technique derived from the product rule of differentiation, used to integrate the product of two functions. The formula is $\int u.vdx=u\int vdx-\int \left[\frac{du}{\mathrm{d}x}.\int vdx\right]$ where u & v are differentiable functions and are commonly designated as first & second function respectively.

1.0What is Integration by Parts?

Integration by parts is derived from the product rule of differentiation. It is used when an integral involves the product of two functions, and one function is easier to differentiate while the other is easier to integrate. The formula for Integration by Parts is:

$\int u.vdx=u\int vdx-\int \left[\frac{du}{\mathrm{d}x}.\int vdx\right]$

Where:

• u is the function you choose to differentiate.
• v is the part you choose to integrate.
• du is the derivative of u.

2.0Integration by Parts Formula (u·v)

To make the integration process easier, you can remember the following formula:

$\int u.vdx=u\int vdx-\int \left[\frac{du}{\mathrm{d}x}.\int vdx\right]$

$=1^{st}function\times integralof{2}^{nd}–\int \left(diff.coeff.of{I}^{st}\right)\times \left(Integralof{2}^{nd}\right)\mathrm{d}x$

3.0Order of Integration by Parts

The order in which you choose u and v can impact how simple the resulting integral is. Following the ILATE rule typically provides the best results, but it may take some practice to recognize the most efficient choice.

4.0ILATE Rule for Integration by Parts

To decide which function should be u (to differentiate), we can use the ILATE rule, which is a helpful mnemonic:

• I: Inverse trigonometric functions (like $\arcsin x$)
• L: Logarithmic functions (like $Inx$)
• A: Algebraic functions (like $x^2$)
• T: Trigonometric functions (like $\sin x$)
• E: Exponential functions (like $e^x$)

According to this rule, choose u from the function that appears first in the list.

5.0Integration by Parts for Definite Integrals

When you’re dealing with definite integrals, the process is similar to indefinite integration, but you must evaluate the integral at the upper and lower limits. This makes the final solution specific to the given range.

6.0Integration by Parts Proof

The proof of the Integration by Parts formula comes from the product rule of differentiation. By differentiating the product of two functions, we get:

Let $I=\int f\left(x\right).g\left(x\right)\mathrm{d}x==f\left(x\right).\int g\left(x\right)dx-\int \left(f^{\prime }\left(x\right)\right)\left(g\left(x\right)\mathrm{d}x\right)dx$

$=1^{st}function\times integralof{2}^{nd}–\int \left(diff.coeff.of{I}^{st}\right)\times \left(Integralof{2}^{nd}\right)\mathrm{d}x$

Proof : $\frac{d}{\mathrm{d}x}\left[f\left(x\right).g\left(x\right)\right]=f\left(x\right).g^{\prime }\left(x\right)+g\left(x\right).f^{\prime }\left(x\right)$

∴ $\int f\left(x\right).g^{\prime }\left(x\right)\mathrm{d}x=f\left(x\right).g\left(x\right)-\int g\left(x\right).f^{\prime }\left(x\right)\mathrm{d}x$

Note :

When applying the integration by parts rule, careful attention must be given to selecting the first and second functions. Generally, the following methods are used:

1. If one of the functions is not easily integrable (such as $Inx,{\sin }^{-1}x,{\cos }^{-1}x,{\tan }^{-1}x,$ etc.), it is chosen as the first function, and the other function is taken as the second function. For example, in the integration of   is taken as the first function, and x as the second.
2. If there is no other function to choose from, unity (1) is taken as the second function. For example, in the integration of $\int {\tan }^{-1}xdx,{\tan }^{-1}x$ is taken as the first function, and 1 as the second.
3. If both functions are directly integrable, the first function is selected so that its derivative simplifies the integral. To make this choice, we follow a preference order: Inverse functions, Logarithmic functions, Algebraic functions, Trigonometric functions, and Exponential functions, abbreviated as ILATE. In this order, the function on the left is always chosen as the first function. For example, in the integration of , x is chosen as the first function, and sin x as the second.

This approach ensures that the resulting integral is simpler and more manageable.

7.0Examples of Integration by Parts with Solutions

Example 1: $\int e^{Inx+x}.dx$

Solution:

$x\int e^{x}dx-\int \left(1\right)\left(e^{xdx}\right)\mathrm{d}x$

$x{e}^x-\int e^{x}dx$

$x{e}^x-e^x+C$

Example 2: $\int x^4\ell nxdx$

Solution:

$\Rightarrow \ell nx\int x^{4}dx-\int \left(\frac{d\left(\ell nx\right)}{\mathrm{d}x}\int x^{4}dx\right)\mathrm{d}x$

$\Rightarrow \ell nx.\frac{x^5}{5}-\int \frac{1}{x}\times \frac{x^5}{5}\mathrm{d}x$

$\Rightarrow \frac{x^5}{5}\ell nx-\frac{x^5}{25}+C$

Example 3: $\int \ell n{\left(2x+3\right)}^{\left(2x+3\right)}\mathrm{d}x$

Solution:

• $\Rightarrow \int \left(2x+3\right)\ell n\left(2x+3\right)\mathrm{d}x$
• $\Rightarrow \ell n\left(2x+3\right)\int \left(2x+3\right)\mathrm{d}x-\int \frac{1}{2x+3}\left(\int \left(2x+3\right)\mathrm{d}x\right)dx$
• $\Rightarrow \ell n\left(2x+3\right)\frac{{\left(2x+3\right)}^2}{4}-\int \frac{1}{2x+3}\times \frac{{\left(2x+3\right)}^2}{2}\mathrm{d}x$
• $\Rightarrow \ell n\left(2x+3\right)\frac{{\left(2x+3\right)}^2}{4}-\frac{{\left(2x+3\right)}^2}{4}+C$

Example 4: $\int {\sin }^{-1}xdx$

Solution:

• Let 1 is algebraic function
• $\int 1.{\sin }^{-1}xdx$
• ${\sin }^{-1}x\left(x\right)-\int \left(\frac{1}{\sqrt{1-x^2}}\right)\left(x\right)\mathrm{d}x$
• $1-x^2=t$
• $-2xdx=dt$
• $x{\sin }^{-1}x+\frac{1}{2}\int \frac{dt}{\sqrt{t}}$
• $x{\sin }^{-1}x+\frac{1}{2}\frac{\sqrt{t}}{\left(\frac{1}{2)}\right)}+C$
• $x{\sin }^{-1}x+\sqrt{1-x^2}+C$

Example 5: Evaluate: $\int \frac{x}{1+\sin x}\mathrm{d}x$

Solution:

Let, $I=\int \frac{x}{1+\sin x}\mathrm{d}x=\int \frac{x\left(1-\sin x\right)}{\left(1+\sin x\right)\left(1-\sin x\right)}$

$=\int \frac{x\left(1-\sin x\right)}{1-{\sin }^{2}x}\mathrm{d}x=\int \frac{x\left(1-\sin x\right)}{{\cos }^{2}x}\mathrm{d}x=\int x{\sec }^{2}xdx-\int x\sec x\tan xdx$

$\left[x\int {\sec }^{2}xdx-\int \left\{\frac{\mathrm{d}x}{\mathrm{d}x}\int {\sec }^{2}xdx\right\}\mathrm{d}x\right]-\left[x\int \sec x\tan xdx-\int \left\{\frac{\mathrm{d}x}{\mathrm{d}x}\int \sec x\tan xdx\right\}\mathrm{d}x\right]$

$=\left[x\tan x-\int \tan xdx\right]-\left[x\sec x-\int \sec xdx\right]$

$=\left[x\tan x-\ell n|secx|\right]-[xsecx-\ell n\left|\sec x+\tan x\right|]+C$

$=x\left(\tan x-\sec x\right)+\ell n\left|\frac{\left(\sec x+\tan x\right)}{\sec x}\right|+C$

Ans. $=\frac{-x\left(1-\sin x\right)}{\cos x}+\ell n\left|1+\sin x\right|+C$

8.0Practice Questions on Integration by Parts

1. $\int xIn\left(x^2+1\right)\mathrm{d}x$
2. $\int e^x\sin xdx$
3. $\int x^2{e}^{x}dx$

4.  What is Integration by Parts?

Integration by parts is a technique used to integrate the product of two functions. It is based on the product rule of differentiation and follows the formula $\int u.vdx=u\int vdx-\int \left[\frac{du}{\mathrm{d}x}.\int vdx\right]$