Mastering Integration by Parts: The Ultimate Guide

Welcome to the ultimate guide on mastering integration by parts. If you’re a student of calculus, you’ve likely encountered integration problems that seem insurmountable. That’s where integration by parts comes to your rescue. In this comprehensive guide, we’ll break down this essential technique step by step, ensuring you gain the confidence and skills needed to tackle complex integrals.

Understanding the Basics

Before diving into integration by parts, let’s ensure we have a solid foundation in the basics. Integration, at its core, is a way of finding the total accumulation of something over an interval. It’s like measuring the area under a curve. But what’s the link between integration and integration by parts? It’s differentiation.

Differentiation is the process of finding the rate of change of a function. In integration by parts, we use the product rule, a concept from differentiation, but in reverse.

Now, let’s introduce the integration by parts formula.

Integration by Parts Formula

The integration by parts formula is a powerful tool that connects two functions in a way that makes integration more manageable. It’s represented as:

$\displaystyle \int u dv = uv-\int v du$

Here’s what each term means:

• $\displaystyle \int$ represents the integral sign.
• $u$ is a function you choose.
• $dv$ is another function you choose.
• $du$ is the derivative of $u$.
• $v$ is the integral of $dv$.

Now, the real magic of integration by parts happens when you carefully select $u$ and $dv$. The key is to pick $u$ in a way that makes du simpler to work with than u itself. It’s an art, and practice will sharpen your skills.

Let’s illustrate this with an example:

Integration by parts isn’t just about solving basic problems. It has advanced applications in calculus and various fields of science. For instance, it’s used in finding moments of inertia, evaluating certain types of definite integrals, and even in probability theory.

Let’s explore a bit further:

Moment of Inertia

In physics and engineering, the moment of inertia describes how objects resist changes in their rotational motion. Integration by parts can help calculate moments of inertia for complex shapes.

Definite Integrals

Integration by parts can simplify definite integrals, making them more manageable. This is especially useful when dealing with improper integrals.

Probability Theory

In probability, integration by parts can help derive probability density functions and calculate expected values.

Pitfalls to Avoid

As you delve into integration by parts, be aware of common pitfalls:

Incorrect $u$ and $dv$ Selection

Choosing the wrong $u$ and $dv$ can lead to more complex integrals instead of simplifying them. Practice is crucial for improving your selection skills.

Algebraic Errors

Simple algebraic mistakes can lead to incorrect results. Always double-check your work, especially when you have multiple steps.

Integration by parts is a technique for evaluating definite integrals or indefinite integrals.

This process is often called Partial Integration and is a process that finds the integral of a product of functions in terms of the integral of the product of their derivative and antiderivative.

It is based on the formula:

\large \displaystyle \begin{align} \int uv^{\prime} = & \ uv-\int u^{\prime}v \\ &\text{or} \\ \int u^{\prime}v = & \ uv-\int uv^{\prime} \end{align}

This formula allows us to rewrite the integral of the product of two functions as the product of the two functions minus the integral of the product of their derivatives.

Here’s an example of how to use the formula to evaluate an integral:

$\displaystyle \int x \sin x dx$

We can choose:

$\begin{array}{ll} u=x, & v^{\prime}=\sin x \\ u^{\prime}=1, & v=-\cos x \end{array}$

Substituting these values into the formula, we get:

\begin {align} \displaystyle \int x \sin x dx &= x (-\cos x)-\int (-\cos x) dx \\ &= -x\cos x+\int \cos x dx \end{align}

The second integral on the right-hand side is easy to evaluate since it is a standard integral:

$\displaystyle \int \cos x dx = \sin x + C$

Substituting this back into the original equation, we get:

$\displaystyle \int x \sin x dx = -x \cos x+\sin x + C$

And that’s it! We’ve successfully evaluated the integral using integration by parts.

Example 1 – Integration by Parts of Trigonometric Functions

(a)     Find $\displaystyle \int x \cos x dx$.

\begin{align} \displaystyle \require{AMSsymbols} &\color{green}{u = x, v^{\prime}=\cos x} \\ &\color{green}{u^{\prime}=1, v=\sin x} \\ \int x \cos x dx &= x \sin x-\int1 \times \sin x dx \\ &= x \sin x + \cos x +C \end{align}

(b)     Hence, find $\displaystyle \int x^2 \sin x dx$.

\begin{align} \require{AMSsymbols} \displaystyle &\color{green}{u = x^2, v^{\prime}= \sin x} \\ &\color{green}{u^{\prime}= 2x, v=-\cos x} \\ \int x^2 \sin x dx &= x^2(-\cos x)-\int 2x(-\cos x) dx \\ &= -x^2 \cos x+2\int x \cos x dx \\ &= -x^2 \cos x + 2 (x \sin x + \cos x) +C &\color{green}{\text{by (a)}} \end{align}

Example 2 – Integration by Parts using Double Angle Formulae

Prove $\displaystyle \int \left( \cos^4 x-\sin^4 x \right) dx = \frac{1}{4} \sin^3 x \cos x-\frac{3}{32} \sin 4x + \frac{3}{16} \sin 2x + \cos^3 x \sin x + C$.

\require{AMSsymbols} \displaystyle \begin{align} \int \cos^4 x dx &= \int \cos^3 x \cos x dx \\ &\color{green}{u=\cos^3 x, v^{\prime}=\cos x} \\ &\color{green}{u^{\prime}=\cos^3 x \sin x, v=\sin x} \\ &= \cos^3 x \sin x-\int-3 \cos^2 x \sin^2 x dx \\ &\color{green}{\int uv^{\prime} = uv-\int u^{\prime}v} \\ &= \cos^3 x \sin x + 3\int \cos^2 x \sin^2 x dx \\ &= \cos^3 x \sin x + 3\int \left(\cos x \sin x \right)^2 dx \\ &= \cos^3 x \sin x + 3\int \left( \frac{\sin 2x}{2} \right)^2 dx \\ &\color{green}{\sin 2x = 2 \cos x \sin x} \\ &= \cos^3 x \sin x + \frac{3}{4} \int \sin^2 2x dx \\ &= \cos^3 x \sin x + \frac{3}{4} \int \frac{1-\cos 4x}{2} dx \\ &\color{green}{\cos 4x = 1-2 \sin^2 x} \\ &= \cos^3 x \sin x + \frac{3}{8} \int \left(1-\cos 4x \right) dx \\ &= \cos^3 x \sin x + \frac{3}{8} \left(x-\frac{\sin 4x}{4} \right)+C_1 \\ &= \cos^3 x \sin x + \frac{3x}{8}-\frac{3}{32} \sin 4x+C_1 \color{green}{\cdots (1)} \\ \int \sin^4 x dx &= \int \sin^3 x \sin x dx \\ &\color{green}{u=\sin^3 x, v^{\prime}= \sin x} \\ &\color{green}{u^{\prime}=3 \sin^2 x \cos x, v=-\cos x} \\ &= \sin^3 x (-\cos x)-\int 3 \sin^2 x \cos x (-\cos x)dx \\ &\color{green}{\int uv^{\prime}=uv-\int u^{\prime}v} \\ &= -\sin^3 x \cos x+3\int \sin^2 x \cos^2 x dx \\ &= -\sin^3 x \cos x+3\int \sin^2 x (1-\sin^2 x) dx \\ \int \sin^4 x dx &= -\sin^3 x \cos x+3\int \sin^2 x dx-3 \int \sin^4 x dx \\ \int \sin^4 x dx + 3\int \sin^4 x dx &= -\sin^3 x \cos x+3\int \sin^2 x dx \\ 4\int \sin^4 x dx &= -\sin^3 x \cos x+3\int \frac{1-\cos 2x}{2} dx \\ &\color{green}{\cos 2x = 1-2 \sin^2 x} \\ &= -\sin^3 x \cos x+\frac{3}{2}\int \left( 1-\cos 2x \right) dx \\ &= -\sin^3 x \cos x+\frac{3}{2} \left( x-\frac{\sin 2x}{2} \right)+C_2 \\ &= -\sin^3 x \cos x+\frac{3x}{2}-\frac{3}{4} \sin 2x +C_2 \\ \int \sin^4 x dx &= -\frac{1}{4}\sin^3 x \cos x+\frac{3x}{8}-\frac{3}{16} \sin 2x + C_2 \color{green}{\cdots (2)} \\ \int \left( \cos^4 x-\sin^4 x \right) dx &= \int \cos^4 x dx-\int \sin^4 x dx \\ &\color{green}{\text{by (1) and (2)}} \\ &= \left( \cos^3 x \sin x + \frac{3x}{8}-\frac{3}{32} \sin 4x +C_1 \right)-\left(-\frac{1}{4}\sin^3 x \cos x+\frac{3x}{8}-\frac{3}{16} \sin 2x+ C_2 \right) \\ \therefore \int \left( \cos^4 x-\sin^4 x \right) dx &= \frac{1}{4} \sin^3 x \cos x-\frac{3}{32} \sin 4x + \frac{3}{16} \sin 2x + \cos^3 x \sin x + C \\ &\color{green}{C=C_1-C_2} \end{align}

Example 3 – Integration by Parts of Exponential Functions

(a)     Find $\displaystyle \int y e^y dy$.

\begin{align} \displaystyle \require{AMSsymbols} &\color{green}{u = y, v^{\prime}=e^y} \\ &\color{green}{u^{\prime}=1, v=e^y} \\ \int y e^y dy &= y e^y-\int e^y dy \\ &= y e^y-e^y +C \end{align}

(b)     Find $\displaystyle \int x^3 e^{x^2} dx$.

\begin{align} \displaystyle \require{AMSsymbols} \text{Let } y &= x^2 \\ \frac{dy}{dx} &= 2x \\ dx &= \frac{dy}{2x} \\ \int x^3 e^{x^2} dx &= \int x^3 e^y \frac{dy}{2x} \\ &= \frac{1}{2} \int x^2 e^y dy \\ &= \frac{1}{2} \int y e^y dy \\ &= \frac{1}{2} \left(y e^y-e^y \right) +C &\color{green}{\text{by (a)}} \\ &= \frac{1}{2} \left( x^2 e^{x^2}-e^{x^2} \right)+C \end{align}

Example 4 – Integration by Parts of Natural Logarithmic Functions

(a)     Find $\displaystyle \int \ln x dx$.

\begin{align} \displaystyle \require{AMSsymbols} &\color{green}{u = \ln x, v^{\prime}=1} \\ &\color{green}{u^{\prime}=\frac{1}{x}, v=x} \\ \int \ln x dx &= \int \ln x \times 1 dx \\ &= \left( \ln x \right) x-\int \frac{1}{x} \times x dx \\ &= x \ln x-\int 1 dx \\ &= x \ln x-x +C \end{align}

(b)     Hence, find $\displaystyle \int (\ln x)^2 dx$.

\begin{align} \displaystyle \require{AMSsymbols} &\color{green}{u = (\ln x)^2, v^{\prime}=1} \\ &\color{green}{u^{\prime}=2 (\ln x) \frac{1}{x}, v=x} \\ \int (\ln x)^2 dx &= \int (\ln x)^2 \times 1 dx \\ &= (\ln x)^2 x-\int 2(\ln x) \frac{1}{x} \times x dx +C \\ &= x (\ln x)^2-2\int \ln x dx +C \\ &= x (\ln x)^2-2 (x \ln x-x)+C &\color{green}{\text{by (a)}} \\ &= x(\ln x)^2-2x \ln x + 2x +C \end{align}

Example 5 – Integration by Parts of Exponential and Trigonometric Functions

(a)     Prove $\displaystyle \int e^x \cos x dx = e^x \sin x-\int e^x \sin x dx$.

\begin{align} \displaystyle \require{AMSsymbols} &\color{green}{u = e^x, v^{\prime}=\cos x} \\ &\color{green}{u^{\prime}=e^x, v=\sin x} \\ \int e^x \cos x dx &= e^x \sin x-\int e^x \sin x dx \end{align}

(b)     Hence, find $\displaystyle \int e^x \cos x dx$.

\begin{align} \require{AMSsymbols} \displaystyle \int e^x \cos x dx &= e^x \sin x-\int e^x \sin x dx \color{green}{\cdots (1)} \\ &\color{green}{u = e^x, v^{\prime}=\sin x} \\ &\color{green}{u^{\prime}=e^x, v=-\cos x} \\ \int e^x \sin x dx &= e^x (-\cos x)-\int e^x (-\cos x)dx \\ &= -e^x \cos x + \int e^x \cos x dx \\ \color{green}{(1) \cdots} \int e^x \cos x dx &= e^x \sin x-\left(-e^x \cos x + \int e^x \cos x dx\right) \\ &=e^x \sin x + e^x \cos x-\int e^x \cos x dx \\ 2 \int e^x \cos x dx &= e^x \sin x + e^x \cos x \\ \therefore \int e^x \cos x dx &= \frac{e^x \sin x + e^x \cos x}{2} + C \end{align}

(c)     Hence, find $\displaystyle \int e^x \sin x dx$.

\begin{align} \require{AMSsymbols} \displaystyle \int e^x \cos x dx &= e^x \sin x-\int e^x \sin x dx &\color{green}{\text{by (a)}} \\ \int e^x \sin x dx &= e^x \sin x-\int e^x \cos x dx \\ &= e^x \sin x-\frac{e^x \sin x + e^x \cos x}{2} &\color{green}{\text{by (b)}} \\ &= \frac{e^x \sin x-e^x \cos x}{2} +C \end{align}

Example 6 – Integration by Parts of Quadratic and Trigonometric Functions

Find $\displaystyle \int x^2 \cos x dx$.

\begin{align} \require{AMSsymbols} \displaystyle &\color{green}{u = x^2, v^{\prime}=\cos x} \\ &\color{green}{u^{\prime}=2x, v=\sin x} \\ \int x^2 \cos x dx &= x^2 \sin x-2 \int x \sin x dx \color{green}{\cdots (1)} \\ \int x \sin x dx &= x (-\cos x)-\int 1 \times (-\cos x) dx \\ &= -x \cos x+\int \cos x dx \\ &= -x \cos x + \sin x \\ \color{green}{(1) \cdots} \int x^2 \cos x dx &= x^2 \sin x-2(-x \cos x+\sin x)+C \end{align}

Example 7 – Integration by Parts of Inverse Sine Functions

(a)     Find $\displaystyle \int \frac{x}{\sqrt{1-x^2}} dx$.

\begin{align} \require{AMSsymbols} \displaystyle \text{Let } u &= 1-x^2 \\ \frac{du}{dx} &=-2x \leadsto \frac{du}{-2x} = dx \\ \int \frac{x}{\sqrt{1-x^2}} dx &= \int \frac{x}{\sqrt{u}} \frac{du}{-2x} \\ &= -\frac{1}{2} \int \frac{1}{\sqrt{u}} du \\ &= -\frac{1}{2} \int u^{-\frac{1}{2}} du \\ &= -\sqrt{u}+C \\ &= -\sqrt{1-x^2}+C \end{align}

(b)     Hence, find $\displaystyle \int \sin^{-1} x dx$.

\begin{align} \require{AMSsymbols} \displaystyle &\color{green}{u^{\prime} = \frac{1}{\sqrt{1-x^2}}, v=x} \\ &\color{green}{u=\sin^{-1} x, v^{\prime}=1} \\ \int \sin^{-1} x dx &= x \sin^{-1} x-\int \frac{x}{\sqrt{1-x^2}} dx \\ &= s \sin^{-1} x-\left(-\sqrt{1-x^2} \right)+C &\color{green}{\text{by (a)}} \\ &= x \sin^{-1} x+\sqrt{1-x^2}+C \end{align}

Example 8 – Integration by Parts of Inverse Tangent Functions

Find $\displaystyle \int \tan^{-1} x dx$.

\begin{align} \require{AMSsymbols} \displaystyle &\color{green}{u = \tan^{-1} x, v^{\prime}=1} \\ &\color{green}{u^{\prime}=\frac{1}{1+x^2}, v=x} \\ \int \tan^{-1} x dx &= x \tan^{-1} x-\int \frac{x}{1+x^2} dx \\ &= x \tan^{-1} x-\frac{1}{2} \int \frac{2x}{1+x^2} dx \\ &= x \tan^{-1} x-\frac{1}{2} \ln \left|1+x^2 \right| + C \end{align}

Example 9 – Integration by Parts of Secant Functions

(a)     Prove $\displaystyle \frac{d}{dx} \sec x = \sec x \tan x$.

\begin{align} \require{AMSsymbols} \displaystyle \frac{d}{dx} \sec x &= \frac{d}{dx} (\cos x)^{-1} \\ &= -(\cos x)^{-2} \times \frac{d}{dx} \cos x \\ &= -\sec^2 x \times (-\sin x) \\ &= \frac{1}{\cos^2 x} \times \sin x \\ &= \frac{1}{\cos x} \times \frac{\sin x}{\cos x} \\ &= \sec x \tan x \end{align}

(b)     Prove $\displaystyle \frac{d}{dx} \tan x = \sec^2 x$.

\begin{align} \require{AMSsymbols} \displaystyle \frac{d}{dx} \tan x &= \frac{d}{dx} \frac{\sin x}{\cos x} \\ &= \frac{(\sin x)^{\prime} \cos x-\sin x (\cos x)^{\prime}}{\cos^2 x} \\ &= \frac{\cos x \cos x-\sin x (-\sin x)}{\cos^2 x} \\ &= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} \\ &= \frac{1}{\cos^2 x} \\ &= \sec^2 x \end{align}

(c)     Hence, find $\displaystyle \int \sec x dx$.

\begin{align} \require{AMSsymbols} \displaystyle \int \sec x dx &= \int \frac{\sec x (\sec x + \tan x)}{\sec x +\tan x} dx \\ &= \int \frac{\sec^2 x + \sec x \tan x}{\sec x +\tan x} dx \\ &= \int \frac{(\tan x)^{\prime} + (\sec x)^{\prime}}{\sec x + \tan x} dx \\ &= \int \frac{(\tan x+\sec x)^{\prime}}{\sec x + \tan x} dx \\ &= \ln \left|\sec x + \tan x \right| + C \end{align}

(d)     Hence, find $\displaystyle \int \sec^3 x dx$.

\begin{align} \require{AMSsymbols} \displaystyle &\color{green}{u = \sec x, v^{\prime}=\sec^2 x} \\ &\color{green}{u^{\prime}=\sec x \tan x, v=\tan x} \\ \int \sec^3 x dx &= \sec x \tan x-\int \sec x (\sec^2 x-1) dx \\ &= \sec x \tan x-\int \sec^3 x dx + \int \sec x dx \\ 2\int \sec^3 x dx &= \sec x \tan x + \int \sec x dx \\ &= \sec x \tan x + \ln \left| \sec x + \tan x \right| &\color{green}{\text{by (c)}} \\ \therefore \int \sec^3 x dx &= \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln \left| \sec x + \tan x \right| + C \end{align}

(e)     Find $\displaystyle \int x \sec x \tan x dx$.

\begin{align} \require{AMSsymbols} \displaystyle &\color{green}{u = x, v^{\prime}= \sec x \tan x} \\ &\color{green}{u^{\prime}= 1, v= \sec x} &\color{green}{\text{by (a)}} \\ \int x \sec x \tan x dx &= x \sec x-\int \sec x dx \\ &= x \sec x-\ln \left| \sec x+\tan x \right|+C &\color{green}{\text{by (c)}} \end{align}

(f)     Find $\displaystyle \int \frac{1}{\sqrt{1+x^2}} dx$ by letting $x=\tan \theta$.

\begin{align} \require{AMSsymbols} \displaystyle dx &= \sec^2 \theta d \theta \\ \int \frac{1}{\sqrt{1+x^2}} dx &= \int \frac{1}{\sqrt{1+\tan^2 \theta}} \sec^2 \theta d \theta \\ &= \int \frac{1}{\sqrt{\sec^2 \theta}} \sec^2 \theta d \theta \\ &= \int \frac{1}{\sec \theta} \sec^2 \theta d \theta \\ &= \int \sec \theta d \theta \\ &= \ln \left| \sec \theta + \tan \theta \right| &\color{green}{\text{by (c)}} \\ &= \ln \left| x+\sqrt{1+x^2} \right|+C &\color{green}{\sec \theta = \sqrt{1+\tan^2 \theta} = \sqrt{1+x^2}} \end{align}

(g)     Hence, find $\displaystyle \int \sqrt{1+x^2} dx$.

\begin{align} \require{AMSsymbols} \displaystyle &\color{green}{u = \sqrt{1+x^2}, v^{\prime}= 1} \\ &\color{green}{u^{\prime}= \frac{x}{\sqrt{1+x^2}}, v= x} \\ \int \sqrt{1+x^2} dx &= x \sqrt{1+x^2}-\int \frac{x^2}{\sqrt{1+x^2}} dx \\ &= x \sqrt{1+x^2}-\int \frac{x^2+1-1}{\sqrt{1+x^2}} dx \\ &= x \sqrt{1+x^2}-\int \sqrt{1+x^2} dx + \int \frac{1}{\sqrt{1+x^2}} dx \\ \int \sqrt{1+x^2} dx &= x \sqrt{1+x^2}-\int \sqrt{1+x^2} dx + \ln \left| x+\sqrt{1+x^2} \right| &\color{green}{\text{by (f)}} \\ 2 \int \sqrt{1+x^2} dx &= x \sqrt{1+x^2} + \ln \left| x+\sqrt{1+x^2} \right| \\ \therefore \int \sqrt{1+x^2} dx &= \frac{x}{2} \sqrt{1+x^2} + \frac{1}{2} \ln \left| x+\sqrt{1+x^2} \right| +C\end{align}

Example 10 – Integration by Parts of Cosine and Square Root Functions

Find $\displaystyle \int \cos \sqrt{x} dx$.

\begin{align} \require{AMSsymbols} \displaystyle \text{Let } \sqrt{x} &= t \\ x &= t^2 \\ dx &= 2t dt \\ \int \cos \sqrt{x} dx &= \int \cos t \ 2t dt \\ &= 2 \int t \cos t dt \color{green}{\cdots (1)} \\ &\color{green}{u = t, v^{\prime}=\cos t} \\ &\color{green}{u^{\prime}=1, v=\sin t} \\ \int t \cos t dt &= t \sin t-\int \sin t dt \\ &= t \sin t + \cos t \\ \color{green}{(1) \cdots} \int \cos \sqrt{x} dx &= 2 \left( t \sin t + \cos t \right) \\ &= 2t \sin t + 2 \cos t \\ &= 2 \sqrt{x} \sin \sqrt{x} + 2 \cos \sqrt{x} +C \end{align}

Example 11 – Integration by Parts of Logarithmic functions

(a)     Find $\displaystyle \int x \ln x dx$.

\begin{align} \require{AMSsymbols} \displaystyle &\color{green}{u^{\prime} = x, v= \ln x} \\ &\color{green}{u= \frac{x^2}{2}, v^{\prime}= \frac{1}{x}} \\ \int x \ln x dx &= \frac{x^2 \ln x}{2}-\int \frac{x}{2} dx \\ &= \frac{x^2 \ln x}{2}-\frac{x^2}{4}+C \end{align}

(b)     Hence, find $\displaystyle \int x (\ln x)^2 dx$.

\begin{align} \require{AMSsymbols} \displaystyle &\color{green}{u^{\prime}= x, v = (\ln x)^2} \\ &\color{green}{u= \frac{x^2}{2}, v^{\prime}= \frac{2 \ln x}{x}} \\ \int x (\ln x)^2 dx &= \frac{x^2 (\ln x)^2}{2}-\int x \ln x dx \\ &= \frac{x^2 (\ln x)^2}{2}-\left(\frac{x^2 \ln x}{2}-\frac{x^2}{4} \right) &\color{green}{\text{by (a)}} \\ &= \frac{x^2 (\ln x)^2}{2}-\frac{x^2 \ln x}{2}+\frac{x^2}{4} +C \end{align}

Example 12 – Integration by Parts of Logarithmic Functions

Find $\displaystyle \int x^2 \ln x dx$.

\begin{align} \require{AMSsymbols} \displaystyle &\color{green}{u^{\prime} = x^2, v= \ln x} \\ &\color{green}{u= \frac{x^3}{3}, v^{\prime}= \frac{1}{x}} \\ \int x^2 \ln x dx &= \frac{x^3 \ln x}{3}-\int \frac{x^2}{3} dx \\ &= \frac{x^3 \ln x}{3}-\frac{x^3}{9}+C \end{align}

Example 13 – Integration by Parts of

Find $\displaystyle \int x \sec^2 x dx$.

\begin{align} \require{AMSsymbols} \displaystyle &\color{green}{u = x, v^{\prime}= \sec^2 x} \\ &\color{green}{u^{\prime}= 1, v= \tan x} \\ \int x \sec^2 x dx &= x \tan x-\int \tan x dx \\ &= x \tan x + \int \frac{-\sin x}{\cos x} dx \\ &= x \tan x + \int \frac{(\cos x)^{\prime}}{\cos x} dx \\ &= x \tan x + \ln \cos x +C \end{align}

Example 14 – Integration by Parts of Secant Functions

Find $\displaystyle \int x \sin^2 x dx$.

\begin{align} \require{AMSsymbols} \displaystyle &\color{green}{u = x, v^{\prime}= \sin^2 x=\frac{1-\cos 2x}{2}} \\ &\color{green}{u^{\prime}= 1, v=\frac{x}{2}-\frac{\sin 2x}{4} } \\ \int x \sin^2 x dx &= x\left(\frac{x}{2}-\frac{\sin 2x}{4} \right)-\int \left( \frac{x}{2}-\frac{\sin 2x}{4} \right) dx \\ &= \frac{x^2}{2}-\frac{x \sin 2x}{4}-\left( \frac{x^2}{4}+\frac{\cos 2x}{8} \right) \\ &= \frac{x^2}{2}-\frac{x \sin 2x}{4}-\frac{x^2}{4}-\frac{\cos 2x}{8} \\ &= \frac{x^2}{4}-\frac{x \sin 2x}{4}-\frac{\cos 2x}{8} + C \end{align}

Example 15 – Integration by Parts of Inverse Sine functions

(a)     Find $\displaystyle \int \frac{x^2}{\sqrt{1-x^2}} dx$ by letting $x=\sin \theta$.

\begin{align} \require{AMSsymbols} \displaystyle x &= \sin \theta \\ dx &= \cos \theta d \theta \\ \int \frac{x^2}{\sqrt{1-x^2}} dx &= \int \frac{\sin^2 \theta}{\sqrt{1-\sin^2 \theta}} \cos \theta d \theta \\ &= \int \frac{\sin^2 \theta}{\cos \theta} \cos \theta d \theta \\ &= \int \sin^2 \theta d \theta \\ &= \int \frac{1-\cos 2 \theta}{2} d \theta \\ &= \frac{\theta}{2}-\frac{\sin 2 \theta}{4} \\ &= \frac{\theta}{2}-\frac{\sin \theta \cos \theta}{2} &\color{green} {\cos^2 \theta = 1-\sin^2 \theta = 1-x^2} \\ &&\color{green}{\cos \theta = \sqrt{1-x^2}} \\ &= \frac{\sin^{-1} x}{2}-\frac{x \sqrt{1-x^2}}{2}+C \end{align}

(b)     Hence, find $\displaystyle \int x \sin^{-1} x dx$.

\begin{align} \require{AMSsymbols} \displaystyle &\color{green}{u^{\prime}= x,v = \sin^{-1} x} \\ &\color{green}{u= \frac{x^2}{2},v^{\prime}= \frac{1}{\sqrt{1-x^2}}} \\ \int x \sin^{-1} x dx &=\frac{x^2 \sin^{-1} x}{2}-\frac{1}{2} \int \frac{x^2}{\sqrt{1-x^2}} dx \\ &= \frac{x^2 \sin^{-1} x}{2}-\frac{1}{2} \left(\frac{\sin^{-1} x}{2}-\frac{x \sqrt{1-x^2}}{2} \right) &\color{green}{\text{by (a)}} \\ &= \frac{x^2 \sin^{-1} x}{2}-\frac{\sin^{-1} x}{4}+\frac{x \sqrt{1-x^2}}{4}+C \end{align}

Example 16 – Integration by Parts of Inverse Tangent Functions

Find $\displaystyle \int x \tan^{-1} x dx$.

\begin{align} \require{AMSsymbols} \displaystyle &\color{green}{u^{\prime}= x,v = \tan^{-1} x} \\ &\color{green}{u= \frac{x^2}{2},v^{\prime}= \frac{1}{1+x^2}} \\ \int x \tan^{-1} x dx &= \frac{x^2 \tan^{-1} x}{2}-\frac{1}{2} \int\frac{x^2}{1+x^2} dx \\ &= \frac{x^2 \tan^{-1} x}{2}-\frac{1}{2} \int \left( 1-\frac{1}{1+x^2} \right) dx \\ &= \frac{x^2 \tan^{-1} x}{2}-\frac{x}{2}+\frac{1}{2} \tan^{-1} x +C \end{align}

Example 17 – Integration by Parts of Square Root Functions

Find $\displaystyle \int x \sqrt{1-x} dx$.

\begin{align} \require{AMSsymbols} \displaystyle &\color{green}{u = x, v^{\prime}= \sqrt{1-x}} \\ &\color{green}{u^{\prime}= 1, v= \frac{-2 \sqrt{(1-x)^3}}{3}} \\ \int x \sqrt{1-x} dx &= x \times \frac{-2 \sqrt{(1-x)^3}}{3}-\int \frac{-2 \sqrt{(1-x)^3}}{3} dx \\ &= -\frac{2x \sqrt{(1-x)^3}}{3}+\frac{2}{3} \int \sqrt{(1-x)^3} dx \\ &= -\frac{2x \sqrt{(1-x)^3}}{3}-\frac{4 \sqrt{(1-x)^5}}{15}+C \end{align}

Example 18 – Integration by Parts of Tangent Square Functions

Find $\displaystyle \int^{\frac{\pi}{4}}_0 x \tan^2 x dx$.

\begin{align} \require{AMSsymbols} \displaystyle &\color{green}{u = x, v^{\prime}= \tan^2 x} \\ &\color{green}{u^{\prime}= 1, v= \tan x-x} \\ \int^{\frac{\pi}{4}}_0 x \tan^2 x dx &= \left[ x(\tan x-x) \right]^{\frac{\pi}{4}}_0-\int^{\frac{\pi}{4}}_0 (\tan x-x) dx \\ &= \left[ x(\tan x-x) \right]^{\frac{\pi}{4}}_0-\int^{\frac{\pi}{4}}_0 \frac{\sin x}{\cos x} dx + \int^{\frac{\pi}{4}}_0 x dx \\ &= \left[ x\tan x-x^2 + \ln \left| \cos x \right| + \frac{x^2}{2} \right]^{\frac{\pi}{4}}_0 \\ &= \left[ x\tan x-\frac{x^2}{2} + \ln \left| \cos x \right| \right]^{\frac{\pi}{4}}_0 \\ &= \frac{\pi}{4}-\frac{\pi^2}{32}-\ln \sqrt{2} \end{align}

Example 19 – Definite Integration by Parts of Inverse Tangent Functions

Find $\displaystyle \int^1_0 x^2 \tan^{-1} x dx$.

\begin{align} \require{AMSsymbols} \displaystyle &\color{green}{u = \tan^{-1} x, v^{\prime}= x^2} \\ &\color{green}{u^{\prime}= \frac{1}{1+x^2}, v= \frac{x^3}{3}} \\ \int^1_0 x^2 \tan^{-1} x dx &= \left[ \frac{x^3 \tan^{-1} x}{3} \right]^1_0-\frac{1}{3} \int^1_0 \frac{x^3}{1+x^2} dx \\ &= \left[ \frac{x^3 \tan^{-1} x}{3} \right]^1_0-\frac{1}{3} \int^1_0 \left( x-\frac{x}{1+x^2} \right) dx \\ &= \left[ \frac{x^3 \tan^{-1} x}{3} \right]^1_0-\frac{1}{3} \left( \frac{x^2}{2}-\frac{1}{2} \int^1_0 \frac{2x}{1+x^2} dx \right) \\ &= \left[ \frac{x^3 \tan^{-1} x}{3} \right]^1_0-\frac{x^2}{6} +\frac{1}{6} \int^1_0 \frac{\left(1+x^2\right)^{\prime}}{1+x^2} dx \\ &= \left[ \frac{x^3 \tan^{-1} x}{3}-\frac{x^2}{6}+\frac{1}{6} \ln \left(1+x^2\right) \right]^1_0 \\ &= \frac{\pi}{12}-\frac{1}{6}+\frac{1}{6} \ln 2 \end{align}

Example 20 – Definite Integration by Parts of Natural Logarithmic Functions

Find $\displaystyle \int^{\infty}_1 \frac{\ln x}{x^2} dx$.

\begin{align} \require{AMSsymbols} \displaystyle &\color{green}{u = \ln x, v^{\prime}= \frac{1}{x^2}} \\ &\color{green}{u^{\prime}= \frac{1}{x}, v=-\frac{1}{x}} \\ \int^{\infty}_1 \frac{\ln x}{x^2} dx &= \left[ -\frac{\ln x}{x} \right]^{\infty}_1 + \int^{\infty}_1 \frac{1}{x^2} dx \\ &= \left[ -\frac{\ln x}{x}-\frac{1}{x} \right]^{\infty}_1 \\ &= 0-(0-1) \\ &= 1 \end{align}

Summary and Conclusion

Congratulations! You’ve completed the ultimate guide to mastering integration by parts. Remember that practice is the key to success in mathematics. As you encounter more integrals, you’ll become more adept at choosing $u$ and $dv$ effectively.

Don’t stop here; continue exploring calculus, and you’ll uncover the beauty and power of mathematics. If you ever find yourself stuck, return to this guide as your trusty reference.

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