Integration by Parts – Definition, Formula & Examples

When you have an integral that is a product of algebraic, exponential, logarithmic, or trigonometric functions, then you can utilise another integration approach called integration by parts. The general rule is to try substitution first, then integrate by parts if that fails. When two functions are multiplied together, with one that can be easily integrated and the other that can be easily separated, integration by parts is typically utilised. In this article, we would learn about integration by parts, its formula and examples. Let’s get this started.

Integration by Parts Definition

The technique of finding the integral of a product of functions in terms of the integral of the product of their derivative and antiderivative is known as integration by parts. It’s typically used to convert the antiderivative of a product of functions into an antiderivative that’s easier to solve. The rule can be thought of as a more comprehensive form of the product differentiation rule.
The goal of integration by parts is to replace a complex integration with a simpler one. It’s typically used to convert the antiderivative of a product of functions into an antiderivative that’s easier to solve. The rule can be thought of as a more comprehensive form of the product differentiation rule.

 Integration by Parts Formula

If \(u\) and \(v\) are any two differentiable functions of a single variable \(x\) (say). Then, by the product rule of differentiation, we have:

\(\frac{d}{d x}(u v)=u \frac{d v}{d x}+v \frac{d u}{d x}\)

Integrating both sides, we get,

\(u v=\int u \frac{d v}{d x} d x+\int v \frac{d u}{d x} d x\)

\(\int u \frac{d v}{d x} d x=u v-\int v \frac{d u}{d x} d x\)

Let \(u=f(x)\) and \(\frac{d v}{d x}=g(x)\). Then,

\(\frac{d u}{d x}=f^{\prime}(x)\) and \(v=\int g(x) d x\)

Therefore, expression \((1)\) can be rewritten as:

\(\int {f\left( x \right)g\left( x \right)dx} \)
\( = f\left( x \right)\int {} g\left( x \right)dx – \int {} \left[ {\int {g\left( x \right)dx} } \right]f’\left( x \right)dx\int {f\left( x \right)g\left( x \right)dx} \)
\( = f\left( x \right)\int {g\left( x \right)dx – \int {} \left[ {f’\left( x \right)\int {g\left( x \right)dx} } \right]dx} \)

If we take \(f\) as the first function and \(g\) as the second function, then this formula may be stated as follows:

The integral of the product of two functions \(=\) (first function) \(\times\) (integral of the second function) \(-\) integral of [(differential coefficient of the first function) \(\times\) (integral of the second function)]

When the product of two functions is presented to us, we use the needed formula, as we learned in integration by parts. By adopting the left term as the first function and the second term as the second function, the integral of the two functions is obtained. The ILATE rule is the name of this approach.

If we need to integrate \(x e^{x}\), we consider \(x\) to be the first function and \(e^{x}\) to be the second. In other words, the initial function is chosen in such a way that the function’s derivative may be simply integrated. This rule’s preference order is usually determined by functions like Inverse trigonometric, Algebraic, Logarithm, Trigonometric, and Exponential. Using the integration by parts formula, we may solve integration by parts examples using this rule.

ILATE RULE: ILATE is a rule that assists in determining which term to distinguish first and which term to integrate first.

\(I=\) Inverse Trigonometric functions

\(L=\) Logarithmic functions

\(A=\) Algebraic functions

\(T=\) Trigonometric functions

\(E=\) Exponential functions

The priority of the first function is like this \(I>L>A>T>E\).

In general, when two functions are multiplied together, and we have to find their integral, then we would take the first function, which is easier to differentiate and the second function, which is easier to integrate.

It is to note that:

1. It’s worth noting that integration by parts isn’t always applicable to the product of functions. For instance, the idea does not work for \(\int \sqrt{x} \sin x d x\). The reason is that there does not exist any function whose derivative is \(\sqrt{x} \sin x\)

2. Notice that we didn’t use any integration constants when calculating the integral of the second function. If we write the integral of the second function \(\cos x\) as \(\sin x+k\), where \(k\) is any constant, then

\(\int {} x\,\cos \,xdx = x\left( {\sin x + k} \right) – \int {} \left( {\sin \,x + k} \right)dx\)
\(= x\left( {\sin \,x + k} \right) – \int {\left( {\sin \,xdx – \int {kdx} } \right)}\)
\(= x\left( {\sin x + k} \right) – \cos \,x – kx + C = x\sin x + \cos \,x + C\)

This shows that adding a constant to the integral of the second function is unnecessary so far as the final result is concerned while applying the method of integration by parts.

3. Usually, if any function is a power of \(x\) or a polynomial in \(x\), then we take it as the first function. However, in cases where other function is inverse trigonometric function or logarithmic function, then we take them as the first function.

Integration by Parts Example

1. Suppose someone asks you to find the integral of,

\(\int x e^{x} d x\)

For this, we can use the integration by parts formula \(\int u v d x=u \int v d x-\int\left[\frac{d}{d x}(u) \int v d x\right] d x\)

From the ILATE rule, we have the first function \(=x\) and the Second function \(=e^{x}\)
Let \(u=x\) and \(v=e^{x}\)

\(I=\int x e^{x} d x=x \int e^{x} d x-\int\left[\frac{d}{d x}(x) \int e^{x} d x\right] d x\)
\( = x\left( {{e^x}} \right) – \int {} \left[ {\left( 1 \right)\left( {{e^x}} \right)} \right]dx\)
\(=x e^{x}-\int e^{x} d x=x e^{x}-e^{x}+C\)

2. Suppose you are asked to find the integral of,

\(\int x \sin 2 x d x\)

In this, we can use the integration by parts formula

\(\int u v d x=u \int v d x-\int\left[\frac{d u}{d x} \int v d x\right] d x\)

From the ILATE rule we have first function \(=x\) and Second function \(=\sin 2 x\)
Let \(u=x\) and \(v=\sin 2 x\)

\(I=\int x \sin 2 x d x=x \int \sin 2 x d x-\int\left[\frac{d}{d x}(x) \int \sin 2 x d x\right] dx\)
\(= x\left( { – \frac{{\cos \,2x}}{2}} \right) – \int {} \left[ {\left( 1 \right)\left( { – \frac{{\cos \,2x}}{2}} \right)} \right]dx =  – x\frac{{\cos \,2x}}{2} + \frac{{\sin \,2x}}{4} + C\)

3. Sometimes we need to use of substitution method before use of the integration of parts in case of

\(\int e^{\sqrt{x}} d x\)

In this example, we first use the substitution method let \(\sqrt{x}=t\)

\(\frac{1}{2 \sqrt{x}} d x=d t d x=2 \sqrt{x} d t d x=2 t d t I=\int e^{\sqrt{x}} d x=\int e^{t} 2 t d x=\int 2 t e^{t} d t\)

Use the integration by parts formula \(\int u v d x=u \int v d x-\int\left[\frac{d}{d x}(u) \int v d x\right] d x\)

Let \(u=t\) and \(v=e^{t}\)

\(I=2\left[t \int e^{t} d t-\int\left[\frac{d[t]}{d t} \int e^{t} d t\right] d t\right]=2\left[t\left(e^{t}\right)-\int(1)\left(e^{t}\right) d t\right]=2\left[t e^{t}-e^{t}\right]+C\)

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

\(I=2\left(\sqrt{x} e^{\sqrt{x}}-e^{\sqrt{x}}\right)+C=2 e^{\sqrt{x}}(\sqrt{x-1})+C\)

4. There is an interesting result which you should remember

\(\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x=e^{x} f(x)+C\)

This can be proved with the help of integration by parts

Proof: We have \(I=\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x=\int e^{x} f(x) d x+\int e^{x} f^{\prime}(x) d x\)

\(=I_{1}+\int e^{x} f^{\prime}(x) d x\), where \(I_{1}=\int e^{x} f(x) d x\)

Taking \(f(x)\) and \(e^{x}\) as the first function and second function, respectively, in \(I\), and integrating it by parts, we have \(I_{1}=f(x) e^{x}-\int f^{\prime}(x) e^{x} d x+C\)

Substituting \(I_{1}\) in above, we get,

\(I=e^{x} f(x)-\int f^{\prime}(x) e^{x} d x+\int e^{x} f^{\prime}(x) d x+C=e^{x} f(x)+C\)

Solved Examples – Integration by Parts

Q.1. Evaluate the following integral:
\(\int x \cos x d x\)
Ans: Use the integration by parts formula
\(\int u v d x=u \int v d x-\int\left[\frac{d}{d x}(u) \int v d x\right] d x\)
Let \(u=x\) and \(v=\cos x\)
\(I = \int {x\cos \,x\,dx = x} \int {\cos \,xdx – \int {\left[ {\frac{d}{{dx}}\left( x \right)\int {\cos \,x\,dx} } \right]dx} } \)
\( = x\left( {\sin x} \right) – \int {} \left[ {\left( 1 \right)\left( {\sin x} \right)} \right]dx\)
\( = x\,\sin \,x – \int {} \sin \,xdx = x\sin x – \left( { – \cos \,x} \right) + C = x\sin x + \cos x + C\)

Q.2. Evaluate the following integral:
\(\int \log (x+1) d x\)
Ans: \(\int \log (x+1) d x=\int(1) \cdot \log (x+1) d x\)
Use the integration by parts formula
\(\int u v d x=u \int v d x-\int\left[\frac{d}{d x}(u) \int v d x\right] d x\)
Let \(u=\log (x+1)\) and \(v=1\)
\(I = \int {\left( 1 \right) \cdot \log \left( {x + 1} \right)dx}  = \log \left( {x + 1} \right)\int {1dx – \int {\left\{ {\frac{d}{{dx}}\left[ {\log \left( {x + 1} \right)} \right]\int {1dx} } \right\}dx} } \)
\(= \log \left( {x + 1} \right) \cdot x – \int {\left\{ {\frac{1}{{x + 1}}\left( x \right)} \right\}dx = x\log \left( {x + 1} \right) – \int {\left( {\frac{x}{{x + 1}}} \right)dx} } \)
\(= x\log \left( {x + 1} \right) – \int {\left( {\frac{{x + 1 – 1}}{{x + 1}}} \right)dx} \)
\(= x\log \left( {x + 1} \right) – \int {\left( {\frac{{x + 1}}{{x + 1}} – \frac{1}{{x + 1}}} \right)dx} \)
\( = x\log \left( {x + 1} \right) – \int {\left( {1 – \frac{1}{{x + 1}}} \right)dx = x\log \left( {x + 1} \right) – x + \log \left( {x + 1} \right) + C} \)
\( = \left( {x + 1} \right)\log \left( {x + 1} \right) – x + C\)

Q.3. Evaluate the following integral:
\(\int x^{2} e^{-x} d x\)
Ans: Use the integration by parts formula
\(\int u v d x=u \int v d x-\int\left[\frac{d}{d x}(u) \int v d x\right] d x\)
Let \(u=x^{2}\) and \(v=e^{-x}\)
\(I = \int {{x^2}{e^{ – x}}dx = {x^2}\int {{e^{ – x}}dx – \int {\left[ {\frac{d}{{dx}}\left( {{x^2}} \right)\int {{e^{ – x}}dx} } \right]dx} } } \)
\( = {x^2}\left( {\frac{{{e^{ – x}}}}{{ – 1}}} \right) – \int {} \left[ {\left( {2x} \right)\left( {\frac{{{e^{ – x}}}}{{ – 1}}} \right)} \right]dx\)
\( =  – {x^2}{e^{ – x}} – \int {\left( { – 2x{e^{ – x}}} \right)dx =  – {x^2}{e^{ – x}} + 2\int {x{e^{ – x}}dx} } \)
Again use the integration by parts formula to evaluate the integration of the second term
Let \(u=x\) and \(v=e^{-x}\)
\(I =  – {x^2}{e^{ – x}} + 2\left[ {x\int {{e^{ – x}}} dx – \int {\left[ {\frac{d}{{dx}}(x)\int {{e^{ – x}}} dx} \right]} dx} \right]\)
\(=  – {x^2}{e^{ – x}} + 2\left[ {x\left( { – {e^{ – x}}} \right) – \int {(1)} \left( { – {e^{ – x}}} \right)dx} \right]\)
\(=  – {x^2}{e^{ – x}} + 2\left[ { – x{e^{ – x}} + \int {{e^{ – x}}} dx} \right]\)
\( =  – {x^2}{e^{ – x}} + 2\left[ { – x{e^{ – x}} – {e^{ – x}}} \right] + C\)
\(=  – {x^2}{e^{ – x}} – 2x{e^{ – x}} – 2{e^{ – x}} + C =  – {e^{ – x}}\left( {{x^2} + 2x + 2} \right) + C\)

Q.4. Evaluate the following integral:
\(\int x^{2} \cos x d x\)
Ans: Use the integration by parts formula
\(\int u v d x=u \int v d x-\int\left[\frac{d u}{d x} \int v d x\right] d x\)
Let \(u=x^{2}\) and \(v=\cos x\)
\(I = \int {{x^2}\cos \,xdx = {x^2}\int {} \cos \,xdx – \int {\left[ {\frac{d}{{dx}}\left( {{x^2}} \right)\int {\cos \,\,xdx} } \right]dx} } \)
\( = {x^2}\left( {\sin \,x} \right) – \int {\left[ {\left( {2x} \right)\left( {\sin \,x} \right)} \right]dx = {x^2}\sin x – 2\int {x\,\sin \,xdx} } \)
Again use the integration by parts formula to evaluate the integration of the second term.
Let \(u=x, v=\sin x\)
\(I = {x^2}\sin \,x – 2\left[ {x\int {} \sin \,xdx – \int {\left[ {\frac{d}{{dx}}\left( x \right)\int {\sin \,xdx} } \right]dx} } \right]\)
\( = {x^2}\sin \,x – 2\left[ {x\left( { – \cos \,x} \right) – \int {\left( 1 \right)\left( { – \cos \,x} \right)dx} } \right]\)
\( = {x^2}\sin x – 2\left[ { – x\cos \,x + \sin \,x} \right] + C = {x^2}\sin \,x + 2x\,\cos \,x – 2\sin \,x + C\)

Q.5. Evaluate the following integral:
\(\int x \tan ^{2} x d x\)
Ans: Use the formula \(\tan ^{2} x=\sec ^{2} x-1\)
\(I=\int x \tan ^{2} x d x=\int x\left(\sec ^{2} x-1\right) d x=\int\left(x \sec ^{2} x-x\right) d x=\int\left(x \sec ^{2} x\right) d x-\int x d x\)
Use the formula by parts formula to evaluate the integration of the first term
Let \(u=x\) and \(v=\sec ^{2} x\) and use the formula
\(\int u v d x=u \int v d x-\int\left[\frac{d}{d x}(u) \int v d x\right] d x \)
\(I = x\int {{{\sec }^2}xdx – \int {\left[ {\frac{d}{{dx}}\left( x \right)\int {{{\sec }^2}xdx} } \right]dx – \frac{{{x^2}}}{2}} } \)
\( = x\left( {\tan \,x} \right) – \int {\left( 1 \right)\tan \,x – \frac{{{x^2}}}{2} = x\tan x – \log \left| {\sec \,x} \right| – \frac{{{x^2}}}{2} + C} \)

Summary

In this article, we have learnt about integration by parts. It is a technique of finding the integral of a product of functions in terms of the integral of the product of their derivative and antiderivative. It’s typically used to convert the antiderivative of a product of functions into an antiderivative that’s easier to solve. When two functions are multiplied together, with one that can be easily integrated and the other that can be easily separated, integration by parts is typically utilised. Integration by parts uses the ILATE rule, which decides the priority of first or second functions.

Frequently Asked Questions (FAQs) – Integration by Parts

Q.1. What is integration by parts?
Ans: The technique of finding the integral of a product of functions in terms of the integral of the product of their derivative and antiderivative is known as integration by parts. It’s typically used to convert the antiderivative of a product of functions into an antiderivative that’s easier to solve. When two functions are multiplied together, with one that can be easily integrated and the other that can be easily separated, Integration by Parts is typically utilised. It uses the ILATE rule, which decides the priority of the first or second functions.

Q.2. What is the idea behind integration by parts?
Ans: When two functions are multiplied together, with one that can be easily integrated and the other that can be easily separated, Integration by Parts is typically utilised. This is done with the help of the ILATE rule, which decides the priority of the first and second functions.

Q.3. What is the integration of 1?
Ans: The integration of \(1\) with respect to \(x\) will be \(x+c\), i.e., \(\int 1 d x=x+c\)

Q.4. What is the product rule of integration?
Ans: The product rule of integration is basically the same as integration by parts. It enables us to integrate the product of two functions. When two functions are multiplied together, with one that can be easily integrated and the other that can be easily separated, the product rule of integration is typically utilised.

Q.5. How do you integrate?
Ans: Integrating a function with respect to x entails calculating the area of the curve with respect to the x-axis. Because integrating is the opposite of differentiating, it is commonly referred to as the antiderivative. Anti-differentiation is equal to integration, according to the fundamental theorem of calculus. In order to integrate any functions, we first need to learn all the standard derivatives formulas as it will help us to find it anti derivative easily.

We hope this detailed article on integration by parts proves helpful to you. If you have any doubts or queries regarding this topic, feel to ask us in the comment section and we will help you at the earliest.