Integration by Parts

Introduction and Formula

In differential calculus, we use the product rule to differentiate the product of two functions. Integration by Parts is the corresponding rule in integral calculus, used to integrate the product of two functions.

Let $u$ and $v$ be differentiable functions of $x$. Then the product rule gives: $$\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$$

Integrating both sides with respect to $x$, we get: $$uv = \int u\frac{dv}{dx}dx + \int v\frac{du}{dx}dx$$

Rearranging, $$\int u\frac{dv}{dx}dx = uv - \int v\frac{du}{dx}dx$$

In differential form, this is written more compactly as: $$\int u \, dv = uv - \int v \, du$$

If we let $u = f(x)$ and $dv = g(x)dx$, then $v = \int g(x)dx$. So the formula may also be written as: $$\int f(x)g(x)dx = f(x)\int g(x)dx - \int \left( f'(x) \int g(x)dx \right)dx$$

In words

The integral of the product of two functions equals:

(first function) $\times$ (integral of second function) $-$ integral of [(derivative of first function) $\times$ (integral of second function)].

Choosing the First and Second Functions (ILATE Rule)

The success of integration by parts depends heavily on making a good choice of the first function $u$ and the second function $dv$. Usually, we choose:

• the first function so that its derivative becomes simpler,
• the second function so that it can be integrated easily.

A standard guideline for choosing the first function is the ILATE (or LIATE) rule. Choose the first function as the one that appears earlier in the following order:

• I: Inverse trigonometric functions (e.g. $\sin^{-1}x, \tan^{-1}x$)
• L: Logarithmic functions (e.g. $\log x, \ln x$)
• A: Algebraic functions (e.g. $x, x^2, x^3+1$)
• T: Trigonometric functions (e.g. $\sin x, \cos x$)
• E: Exponential functions (e.g. $e^x, a^x$)

So, in a product like $x e^x$, we choose $u = x$ and $dv = e^x dx$ because Algebraic comes before Exponential.

This rule is not a rigid law, but it is an extremely useful exam-time guide.

Integrating Logarithmic and Inverse Trigonometric Functions

There are no direct standard formulas for integrals like: $$\int \ln x \, dx, \qquad \int \sin^{-1}x \, dx, \qquad \int \tan^{-1}x \, dx$$

To integrate such functions, we rewrite them as a product with $1$: $$\int \ln x \, dx = \int (\ln x \cdot 1) \, dx$$

Then we apply Integration by Parts.

For example, in $\int \ln x \, dx$:

• choose $u = \ln x$ by the ILATE rule,
• choose $dv = 1\,dx$.

This idea is very important, because it allows us to integrate functions that are not directly in standard anti-derivative form.

Integrals of the form $\int e^x [f(x) + f'(x)] dx$

This is a very important and highly testable special case of integration by parts.

If the integrand is of the form: $$\int e^x [f(x) + f'(x)] dx$$ then the result is immediately: $$\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$$

Why is this true? Because if we differentiate $e^x f(x)$ using the product rule, we get: $$\frac{d}{dx}(e^x f(x)) = e^x f(x) + e^x f'(x) = e^x [f(x) + f'(x)]$$

So the integral is simply the reverse of this derivative.

In many problems, the integrand is not visibly in this form at first. In such cases, you often need to simplify, split, or manipulate the expression so that it becomes $f(x) + f'(x)$ inside the bracket.

Example 1: Basic Integration by Parts

Evaluate: $\int x \cos x \, dx$

Solution: Step 1: Choose $u$ and $dv$ using the ILATE rule. Here, $x$ is Algebraic and $\cos x$ is Trigonometric. So we take: $$u = x, \qquad dv = \cos x \, dx$$

Step 2: Differentiate and integrate. $$du = dx, \qquad v = \int \cos x \, dx = \sin x$$

Step 3: Apply the formula $\int u \, dv = uv - \int v \, du$. $$\int x \cos x \, dx = x \sin x - \int \sin x \, dx$$

Step 4: Evaluate the remaining integral. Since $$\int \sin x \, dx = -\cos x,$$ we get: $$x \sin x - (-\cos x) = x \sin x + \cos x + C$$

Answer: $x \sin x + \cos x + C$

Example 2: Integrating a Logarithmic Function

Evaluate: $\int \ln x \, dx$

Solution: Step 1: Rewrite the integrand as a product: $$\int \ln x \, dx = \int (\ln x \cdot 1) \, dx$$

Step 2: Choose functions. By ILATE, Logarithmic comes before Algebraic, so take: $$u = \ln x, \qquad dv = 1\,dx$$

Step 3: Differentiate and integrate. $$du = \frac{1}{x}dx, \qquad v = \int 1\,dx = x$$

Step 4: Apply integration by parts. $$\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} \, dx$$

Step 5: Simplify the remaining integral. $$= x \ln x - \int 1 \, dx = x \ln x - x + C$$

Step 6: Write neatly. $$= x(\ln x - 1) + C$$

Answer: $x(\ln x - 1) + C$

Example 3: Repeated Integration by Parts

Evaluate: $\int x^2 e^x \, dx$

Solution: Step 1: Choose: $$u = x^2, \qquad dv = e^x dx$$ Then, $$du = 2x\,dx, \qquad v = e^x$$

Step 2: Apply integration by parts. $$\int x^2 e^x dx = x^2 e^x - \int 2x e^x dx$$ $$= x^2 e^x - 2\int x e^x dx$$

Step 3: The remaining integral still needs integration by parts. For $\int x e^x dx$, choose: $$u = x, \qquad dv = e^x dx$$ Then, $$du = dx, \qquad v = e^x$$

Step 4: Apply integration by parts again. $$\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x$$

Step 5: Substitute back into the original expression. $$\int x^2 e^x dx = x^2 e^x - 2(x e^x - e^x) + C$$ $$= x^2 e^x - 2x e^x + 2e^x + C$$

Step 6: Factor out $e^x$. $$= e^x(x^2 - 2x + 2) + C$$

Answer: $e^x(x^2 - 2x + 2) + C$

Example 4: Cyclic Integrals

Evaluate: $I = \int e^x \sin x \, dx$

Solution: Step 1: Choose: $$u = \sin x, \qquad dv = e^x dx$$ Then, $$du = \cos x\,dx, \qquad v = e^x$$

Step 2: Apply integration by parts. $$I = e^x \sin x - \int e^x \cos x \, dx$$

Step 3: Now apply integration by parts to the new integral: Let $$J = \int e^x \cos x \, dx$$ Choose: $$u = \cos x, \qquad dv = e^x dx$$ Then, $$du = -\sin x\,dx, \qquad v = e^x$$

So, $$J = e^x \cos x - \int e^x(-\sin x)dx$$ $$J = e^x \cos x + \int e^x \sin x dx$$ $$J = e^x \cos x + I$$

Step 4: Substitute this into the expression for $I$. $$I = e^x \sin x - J = e^x \sin x - (e^x \cos x + I)$$ $$I = e^x \sin x - e^x \cos x - I$$

Step 5: Bring $I$ to the left side. $$2I = e^x(\sin x - \cos x)$$

Step 6: Divide by 2. $$I = \frac{e^x(\sin x - \cos x)}{2} + C$$

Answer: $\frac{e^x(\sin x - \cos x)}{2} + C$

Example 5: Integrating Inverse Trigonometric Functions

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

Solution: Step 1: Choose functions using ILATE. Inverse trigonometric comes before algebraic, so take: $$u = \tan^{-1}x, \qquad dv = x\,dx$$

Step 2: Differentiate and integrate. $$du = \frac{1}{1+x^2}dx, \qquad v = \int x\,dx = \frac{x^2}{2}$$

Step 3: Apply integration by parts. $$\int x\tan^{-1}x\,dx = \frac{x^2}{2}\tan^{-1}x - \int \frac{x^2}{2(1+x^2)}dx$$ $$= \frac{x^2}{2}\tan^{-1}x - \frac{1}{2}\int \frac{x^2}{1+x^2}dx$$

Step 4: Simplify the fraction inside the integral. Write: $$\frac{x^2}{1+x^2} = \frac{x^2+1-1}{1+x^2} = 1 - \frac{1}{1+x^2}$$

So, $$\int \frac{x^2}{1+x^2}dx = \int \left(1 - \frac{1}{1+x^2}\right)dx$$ $$= x - \tan^{-1}x$$

Step 5: Substitute back. $$\int x\tan^{-1}x\,dx = \frac{x^2}{2}\tan^{-1}x - \frac{1}{2}(x - \tan^{-1}x) + C$$ $$= \frac{x^2}{2}\tan^{-1}x - \frac{x}{2} + \frac{1}{2}\tan^{-1}x + C$$

Step 6: Combine the inverse tangent terms. $$= \frac{1}{2}(x^2+1)\tan^{-1}x - \frac{x}{2} + C$$

Answer: $\frac{1}{2}(x^2 + 1)\tan^{-1} x - \frac{x}{2} + C$

Example 6: Special Form $\int e^x [f(x) + f'(x)] dx$

Evaluate: $\int e^x (\sin x + \cos x) \, dx$

Solution: Step 1: Compare the bracket with the form $f(x) + f'(x)$. Choose: $$f(x) = \sin x$$ Then, $$f'(x) = \cos x$$

Step 2: So the integrand becomes exactly: $$e^x[f(x) + f'(x)]$$

Step 3: Apply the standard result: $$\int e^x[f(x) + f'(x)]dx = e^x f(x) + C$$

Step 4: Substitute $f(x)=\sin x$. $$= e^x \sin x + C$$

Answer: $e^x \sin x + C$

Example 7: Manipulating into the Special Form

Evaluate: $\int \frac{x e^x}{(1+x)^2} dx$

Solution: Step 1: Rewrite the integral as: $$\int e^x \left(\frac{x}{(1+x)^2}\right) dx$$

Step 2: Try to rewrite the bracket in the form $f(x)+f'(x)$. Add and subtract 1 in the numerator: $$x = (x+1)-1$$ So, $$\frac{x}{(1+x)^2} = \frac{x+1-1}{(x+1)^2} = \frac{x+1}{(x+1)^2} - \frac{1}{(x+1)^2}$$ $$= \frac{1}{x+1} - \frac{1}{(x+1)^2}$$

Step 3: Let $$f(x)=\frac{1}{x+1}$$ Then, $$f'(x)= -\frac{1}{(x+1)^2}$$

Step 4: Therefore the integrand becomes: $$e^x[f(x)+f'(x)]$$

Step 5: Apply the theorem. $$\int e^x[f(x)+f'(x)]dx = e^x f(x) + C$$ $$= e^x\left(\frac{1}{x+1}\right) + C$$

Answer: $\frac{e^x}{x+1} + C$

Example 8: Trigonometric Manipulation into Special Form

Evaluate: $\int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx$

Solution: Step 1: Simplify the trigonometric ratio. Using half-angle identities: $$1-\cos x = 2\sin^2\left(\frac{x}{2}\right), \qquad \sin x = 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$

So, $$\frac{1-\sin x}{1-\cos x} = \frac{1-2\sin(x/2)\cos(x/2)}{2\sin^2(x/2)}$$

Step 2: Split the fraction. $$= \frac{1}{2\sin^2(x/2)} - \frac{2\sin(x/2)\cos(x/2)}{2\sin^2(x/2)}$$ $$= \frac{1}{2}\csc^2\left(\frac{x}{2}\right) - \cot\left(\frac{x}{2}\right)$$

Step 3: Rearrange it as $f(x)+f'(x)$. Let $$f(x) = -\cot\left(\frac{x}{2}\right)$$ Then, $$f'(x) = -\left[-\csc^2\left(\frac{x}{2}\right)\cdot \frac{1}{2}\right] = \frac{1}{2}\csc^2\left(\frac{x}{2}\right)$$

Thus the bracket becomes exactly: $$f(x)+f'(x)$$

Step 4: Apply the theorem. $$\int e^x[f(x)+f'(x)]dx = e^x f(x) + C$$ $$= -e^x \cot\left(\frac{x}{2}\right) + C$$

Answer: $-e^x \cot(x/2) + C$