Integration by Parts

Integration by Parts: When you have two differentiable functions of the same variable then, the integral of the product of two functions = (first function) × (integral of the second function) – Integral of [(differential coefficient of the first function) × (integral of the second function)]. This rule is known as integration by parts.

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Integration by Parts

Let’s say that u and v are any two differentiable functions of a single variable x. Then, by the product rule of differentiation, we have

d/dx (uv) = u dv/dx + v du/dx

If we integrate both sides, we get
uv = ∫ [u dv/dx] dx + ∫ [v du/dx] dx
Or, ∫ [u dv/dx] dx = uv – ∫ [v du/dx] dx … (1)

Now, let u = f(x) and dv/dx = g(x). Hence we have
du/dx = f’(x) and v = ∫ g(x) dx

Therefore, equation (1) can be written as
∫ f(x) g(x) dx = f(x) ∫ g(x) dx – ∫ [∫ g(x) dx] f’(x) dx
Or, ∫ f(x) g(x) dx = f(x) ∫ g(x) dx – ∫ [f’(x) ∫ g(x) dx] dx

Integral of the product of two functions (∫ f(x) g(x) dx) = first function [f(x)] x integral of the second function [∫ g(x) dx] – integral of {differential coefficient of the first function [f’(x)] x integral of the second function [∫ g(x) dx] dx]}. This is integration by parts. Let’s look at some examples to understand this well.

Example 1: Find ∫ x cos x dx

Solution: Let, The first function = f(x) = x and the second function = g(x) = cos x. Therefore, according to integration by parts, we have

∫ x cos x dx = x ∫ cos x dx – ∫ [(dx/dt) ∫ cos x dx] dx = x sin x – ∫ sin x dx
= x sin x + cos x + C.

Let’s try the other way round. Let, the first function = f(x) = cos x and the second function = g(x) = x. Therefore,
∫ x cos x dx = cos x ∫ x dx – ∫ {[d(cos x)/ dx] ∫ x } dx
= (cos x) (x²/2) + ∫ (sin x) (x²/2) dx

This is a really complex integral having a higher power of x. Hence, it is necessary to select the first and second function properly.

Browse more Topics under Integrals

• Fundamental Theorem of Calculus
• Introduction to Integration
• Properties of Indefinite Integrals
• Properties of Definite Integrals
• Definite Integral as a Limit of a Sum
• Integration by Partial Fractions
• Integration by Substitutions
• Integral of Some Particular Functions
• Integral of the Type e^x[f(x) + f'(x)]dx

Points to remember

• Integration by parts is not applicable to all functions. For example, it does not work for ∫ √x sin x dx. The reason is that there is no function whose derivative is √x sin x.
• Do not add the constant of integration while finding the integral of the second function. The same can be added once at the end of the integration process. Even if you add it won’t change the answer but simply make the calculation more complex.
• 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 another function is an inverse trigonometric function or a logarithmic function, then we take them as the first function.

Example 2: Find ∫ log x dx

Solution: Do you know any function whose derivative is log x? Guessing it is difficult. Hence, let’s take the first function f(x) = log x. So, the second function g(x) = 1. And, we know that ∫ 1 dx = x. Therefore, ∫ g(x) dx = x. Therefore, we have,

∫ (log x.1) dx = log x ∫ 1 dx – ∫ {[d(log x)/dx] ∫ 1 dx} dx
= x log x – ∫ (1/x) x dx = x log x – ∫ 1 dx
= x log x – x + C

Example 3: Find ∫ x e^x dx

Solution: We know that, ∫ e^x dx = e^x. Hence, we take the first function = f(x) = x and the second function = g(x) = e^x. Therefore, we have,

∫ x e^x dx = x ∫ e^x dx – ∫ [(dx/dx) ∫ e^x dx] dx = x e^x – ∫ 1.e^x dx
= xe^x – e^x + C

Example 4: Find ∫ (x sin^–1 x) / [√(1 – x²)] dx

Solution: Let, The first function = f(x) = sin^–1 x and the second function = g(x) = x / √(1 – x²). Now let’s find ∫ g(x) dx. Let’s substitute (1 – x²) = t, so that dt = – 2x dx or dx = –1/2 dt. Therefore,

∫ [x / √(1 – x²)] dx = –1/2 ∫ dt / √t = – √t = – √(1 – x²)

Hence,
∫ (x sin^–1 x) / [√(1 – x²)] dx = (sin^–1 x) ∫ [x / √(1 – x²)] dx – ∫ {d(sin^–1 x) / dx ∫ [x / √(1 – x²)] dx} dx
= (sin^–1 x) [–√(1 – x²)] – ∫ {[1 / √(1 – x²)] [–√(1 – x²)]} dx … since d(sin^–1 x) / dx = 1 / √(1 – x²).
Simplifying it further, we get,
∫ (x sin^–1 x) / [√(1 – x²)] dx = – √(1 – x²) (sin^–1 x) + ∫ 1.dx
= – √(1 – x²) (sin^–1 x) + x + C.

Example 5: Find ∫ e^x sin x dx

Solution: Let, The first function = f(x) = e^x and the second function = g(x) = sin x. Therefore,
∫ e^x sin x dx = e^x ∫ sin x dx – ∫ [de^x/dx ∫ sin x dx] dx
= ex (– cos x) + ∫ e^x cos x dx …(5.1)

Now, let’s find ∫ e^x cos x dx. For this, let the first function = f₁(x) = e^x and the second function = g₁(x) = cos x. Therefore,
∫ e^x cos x dx = e^x ∫ cos x dx – ∫ [de^x/dx ∫ cos x dx] dx
= e^x sin x – ∫ e^x sin x dx … (5.2)

Substituting (5.2) in (5.1), we get
∫ e^x sin x dx = – e^x cos x + e^x sin x – ∫ e^x sin x dx
Or, 2 ∫ e^x sin x dx = e^x sin x – e^x cos x
Hence, ∫ e^x sin x dx = ½ e^x (sin x – cos x) + C.

More Solved Examples for You

Question: Find ∫ x tan^–1 x

Solution: Let, The first function = f(x) = tan^–1 x and the second function = g(x) = x. Therefore,
I = ∫ x tan^–1 x = tan^–1 x ∫ x dx – ∫ [d(tan^–1 x)/ dx ∫ x dx] dx

Now, we know that the derivative of tan^–1 x with respect to x is 1 / (1 + x²). Also, the integral of x with respect to x is x²/2. Hence, we have
I = (tan^–1 x) (x²/2) – ∫ [1 / (1 + x²)] (x²/2) dx
= (x²/2) tan^–1 x – ½ ∫ x² / (x² + 1) dx = (x²/2) tan^–1 x – ½ ∫ (x² + 1 – 1) / (x² + 1) dx
Hence, I = (x²/2) tan^–1 x – ½ {∫ (x² + 1) / (x² + 1) dx – ∫ dx / (x² + 1)]
So, I = (x²/2) tan^–1 x – ½ [∫dx – ∫ dx / (x² + 1)]

Now, we know that ∫ dx / (x² + a²) = 1/a tan^–1 (x/a) + C. Therefore,
I = (x²/2) tan^–1 x – ½ x + ½ [tan^–1 x] + C
= (x²/2) tan^–1 x – x/2 + ½ (tan^–1 x) + C

Question- What is the product rule of integration?

Answer- The product rule of integration allows one to integrate the product of two functions. For instance, you can turn an equation into a formula useful for integrating through a series of mathematical somersaults.

Question- What is the meaning of integration?

Answer- Integration happens when different people or objects are brought together. For instance, integration of teachers from all of the city’s elementary schools at the middle school.

Question- Who discovered integration by parts?

Answer- Brook Taylor was the one who discovered integration by parts. She was a mathematician who first published the idea of integration by parts in the year 1715.

Question- What is integration by parts used for?

Answer- Integration by Parts refers to a special method of integration that is frequently useful when we multiply two functions together, but it also helps us in other ways too. However, it does not apply to all functions. For instance, it does not work for ∫ √x sin x dx. Therefore, we see that the reason behind is that there is no function whose derivative is √x sin x.