Reduction Formulas For Integration by Parts With Solved Examples

Introduction to Reduction Formulas

Reduction formulas are like the shortcuts of integration - instead of wrestling with higher powers of functions, you break them down step-by-step into easier, lower-order integrals. This technique is incredibly useful in Class 12 calculus, JEE preparation, and any situation where integrals repeat in patterns.

Ever looked at an integral with a huge power and thought, "There has to be an easier way"?

That's exactly where reduction formulas help. They reduce the power of the function step by step until the integral becomes easy to evaluate.

What Is a Reduction Formula?

A reduction formula expresses the integral of a function raised to a power n in terms of an integral with a lower power such as n−1 or n−2.
This repeated reduction continues until the expression becomes a basic integral that can be evaluated directly.

Why do reduction formulas seem difficult when you first learn them?

Most students try to memorize the formulas without understanding where they come from. Once you learn that they are derived using integration by parts, the logic becomes much easier to follow.

Deriving a Standard Reduction Formula

Example: Derivation for

$ \int x^n e^x , dx $

Using Integration by Parts:
Choose

• $u = x^n$

• $dv = e^x dx$

Then
$du = n x^{n-1} dx$ and $v = e^x$

Applying the formula
$ \int u , dv = uv - \int v , du $

We get:
$ \int x^n e^x dx = x^n e^x - n \int x^{n-1} e^x dx $

This is the required reduction formula, expressing the integral in terms of a lower power.

Can reduction formulas really save time in exams?

Absolutely. A problem that might require multiple rounds of integration by parts can often be solved much faster using a reduction formula, making it a valuable tool in competitive examinations.

Reduction Formulas for Trigonometric Integrals

Trigonometric reduction formulas are incredibly useful when you're dealing with higher powers of sine, cosine, tangent, or other trig functions. Instead of integrating directly - which can get messy - these formulas gradually lower the power, making the integral easier step by step. They’re a favorite in competitive exams because they turn long, intimidating expressions into manageable recursive patterns.

Can I derive a reduction formula instead of memorizing it?

Yes. Most reduction formulas are obtained using integration by parts. Understanding the derivation helps you remember the formula naturally and apply it correctly.

Reduction Formula for $ \int \sin^n x , dx $

Let
$ I_n = \int \sin^n x , dx $

Start by rewriting the integrand:
$ I_n = \int \sin^{n-1} x \cdot \sin x , dx $

Using integration by parts and simplifying, you arrive at the classic reduction formula:
$ I_n = -\frac{\sin^{n-1} x \cos x}{n} + \frac{n-1}{n} I_{n-2} $

This means every application drops the power by 2, making it ideal when $n$ is large or even.

Reduction Formula for $ \int \tan^n x , dx $

Let
$ I_n = \int \tan^n x , dx $

Use the identity:
$ \tan^n x = \tan^{n-2} x (\sec^2 x - 1) $

Set $ t = \tan x $, so $ dt = \sec^2 x , dx $.
This transforms a part of the integral into
$ \int \tan^{n-2} x \sec^2 x , dx = \int t^{n-2} , dt $

Finally, the reduction formula becomes:
$ I_n = \frac{\tan^{n-1} x}{n-1} - I_{n-2} $

Example: Solve $ \int \tan^4 x , dx $
You repeatedly apply the reduction formula until the powers drop low enough to integrate directly.

Why do powers like $\sin^n x$ and $\cos^n x$ appear so often in reduction formulas?

These functions follow recurring integration patterns. Reduction formulas take advantage of these patterns and create a systematic method for evaluating higher powers.

Additional Identities for Simplification

Many trig powers can be simplified before applying reduction formulas using identities like:
$ \sin^4 x = \left(\frac{1 - \cos 2x}{2}\right)^2 $
$ \cos^4 x = \left(\frac{1 + \cos 2x}{2}\right)^2 $

These expansions often reduce the complexity before integration even starts.

Reduction Formula for $ \int \cos^n x , dx $

Let
$ I_n = \int \cos^n x , dx $

Using integration by parts, you get:
$ I_n = \frac{\cos^{n-1} x \sin x}{n} + \frac{n-1}{n} I_{n-2} $

This follows the same pattern as the sine reduction formula - reducing the power by 2 on each iteration.

Reduction Formula for $ \int \cot^n x , dx $

Let
$ I_n = \int \cot^n x , dx $

Use the identity:
$ \cot^2 x = \csc^2 x - 1 $

Choose the substitution $ t = \cot x $, then $ dt = -\csc^2 x , dx $.
After simplification, the formula becomes:
$ I_n = -\frac{\cot^{n-1} x}{n-1} - I_{n-2} $

This stepwise reduction makes even large powers of cot manageable.

Reduction Formula for $ \int \sec^n x , dx $

Let
$ I_n = \int \sec^n x , dx $

Using integration by parts and trig identities, the reduction formula is:
$ I_n = \frac{\sec^{n-2} x \tan x}{n-1} + \frac{n-2}{n-1} I_{n-2} $

This is especially helpful when dealing with even powers of secant.

Reduction Formula for $ \int \csc^n x , dx $

Let
$ I_n = \int \csc^n x , dx $

Following a similar process as secant integrals:
$ I_n = -\frac{\csc^{n-2} x \cot x}{n-1} + \frac{n-2}{n-1} I_{n-2} $

How can I tell whether a question requires a reduction formula?

If the integral contains a large power such as $\sin^8 x$, $\cos^7 x$, $x^6e^x$, or $(\log x)^5$, a reduction formula is often the most efficient method.

Advanced Reduction Formulas: Mixed Trigonometric Integrals

Mixed trigonometric integrals often show up when you're working with oscillatory functions or solving physics problems involving waves. A general case you’ll encounter is
$ I_{m,n} = \int \cos^m x \sin nx , dx $,
where m and n can be any positive integers. These integrals look messy at first, but reduction formulas help you shrink them step-by-step into simpler versions.

To derive the reduction, use integration by parts by choosing

• $u = \cos^m x$

• $dv = \sin nxdx$

This choice makes the structure manageable and leads to the following reduction formula:

$ I_{m,n} = -\frac{\cos^m x \cos nx}{m+n} + \frac{m}{m+n} I_{m-1,n-1} $

This result is powerful because it reduces both m and n by 1 at each step. Over repeated applications, the integral eventually collapses into a basic sine-cosine form that can be handled easily.

Additional Useful Results

1. $ \int \cos^m x \cos nx dx = \frac{\cos^m x \sin nx}{m+n} + \frac{m}{m+n} \int \cos^{m-1} x \cos(n-1)x dx $

2. $ \int \sin^m x \sin nx dx = \frac{n \sin^m x \cos nx}{m^2-n^2} - \frac{m \sin^{m-1}x \cos x \cos nx}{m^2-n^2} + \frac{m(m-1)}{m^2-n^2}\int \sin^{m-2}x \sin nx dx $

3. $ \int \sin^m x \cos nx dx = \frac{n \sin^m x \sin nx}{m^2-n^2} - \frac{m \sin^{m-1}x \cos x \cos nx}{m^2-n^2} + \frac{m(m-1)}{m^2-n^2}\int \sin^{m-2}x \cos nx dx $

How to Use Reduction Formulas?

Reduction formulas provide a systematic approach to solving complicated integrals by expressing them in terms of simpler integrals of lower order. Instead of evaluating a difficult integral directly, the reduction formula gradually reduces its complexity until a standard integral is obtained. This method is particularly useful for higher powers of trigonometric, logarithmic, exponential, and algebraic functions.

Step-by-Step Procedure

Using a reduction formula becomes straightforward once you understand the process.

Step 1: Identify the Integral

First, examine the integral and determine its type.

Examples:

• $\int \sin^6 xdx$
• $\int x^5 e^xdx$
• $\int (\log x)^4dx$

These integrals involve higher powers and are suitable candidates for reduction formulas.

Step 2: Select the Appropriate Reduction Formula

Different types of integrals have different reduction formulas.

For example:

• $\int \sin^n xdx$ uses the reduction formula for sine powers.
• $\int \cos^n xdx$ uses the reduction formula for cosine powers.
• $\int x^n e^xdx$ uses the reduction formula for exponential functions.

Choosing the correct formula is crucial for obtaining the right result.

Step 3: Substitute the Value of $n$

Replace the variable $n$ in the reduction formula with the given power.

For example, if the integral is:

$\int \sin^6 xdx$

then substitute:

$n=6$

into the reduction formula.

Step 4: Simplify the Integral

After substitution, simplify the resulting expression.

The formula will produce a new integral with a smaller power.

For example:

$\int \sin^6 xdx$

may be reduced to an expression involving:

$\int \sin^4 xdx$

which is easier to evaluate.

Step 5: Repeat the Process

Apply the reduction formula repeatedly until a basic integral is obtained.

For instance:

$\int \sin^6 xdx$

$\rightarrow \int \sin^4 xdx$

$\rightarrow \int \sin^2 xdx$

$\rightarrow \int 1dx$

Each step reduces the complexity of the problem.

Step 6: Evaluate the Remaining Integral

Once the power becomes sufficiently small, evaluate the remaining integral directly and substitute back to obtain the final answer.

Choosing the Correct Reduction Formula

One of the most important steps is selecting the appropriate reduction formula.

For Trigonometric Functions

Use reduction formulas such as:

• $\int \sin^n xdx$
• $\int \cos^n xdx$
• $\int \tan^n xdx$
• $\int \cot^n xdx$

These formulas help reduce powers of trigonometric functions.

For Exponential Functions

Use formulas involving:

• $\int x^n e^xdx$
• $\int x^n a^xdx$

These commonly appear in calculus and engineering mathematics.

For Logarithmic Functions

Use reduction formulas for:

• $\int (\log x)^ndx$
• $\int x^n \log xdx$

Such integrals are often solved using repeated integration by parts.

For Inverse Trigonometric Functions

Use special reduction formulas for:

• $\int (\sin^{-1}x)^ndx$
• $\int (\tan^{-1}x)^ndx$

These appear in advanced calculus problems.

Quick Selection Guide

Reducing Higher Powers

The primary purpose of a reduction formula is to decrease higher powers step by step.

Example 1: Trigonometric Integral

Consider:

$\int \sin^8 xdx$

The reduction formula reduces the power from:

$8 \rightarrow 6 \rightarrow 4 \rightarrow 2 \rightarrow 0$

At each stage, the integral becomes simpler.

Example 2: Exponential Integral

Consider:

$\int x^5 e^xdx$

Using the reduction formula:

$\int x^5 e^xdx$

is expressed in terms of:

$\int x^4 e^xdx$

which is then reduced further to:

$\int x^3 e^xdx$

and so on.

Example 3: Logarithmic Integral

Consider:

$\int (\log x)^4dx$

The reduction formula successively reduces the power:

$4 \rightarrow 3 \rightarrow 2 \rightarrow 1$

until a basic logarithmic integral remains.

Why Reducing Powers Helps

Higher powers usually make direct integration difficult.

Reduction formulas:

• Simplify lengthy calculations.
• Create a predictable solution pattern.
• Reduce chances of errors.
• Make complicated integrals manageable.

Common Mistakes to Avoid

Students often make avoidable mistakes while applying reduction formulas. Being aware of these errors can improve accuracy significantly.

Using the Wrong Reduction Formula

Always verify the type of integral before selecting a formula.

For example:

• Do not use a cosine reduction formula for a sine integral.
• Do not use a logarithmic reduction formula for an exponential integral.

Substituting the Wrong Value of $n$

Carefully identify the power in the integral.

For:

$\int \sin^7 xdx$

the value is:

$n=7$

not $n=6$.

A small substitution error can affect the entire solution.

Skipping Intermediate Steps

Many students try to jump directly to the final answer.

It is safer to show each reduction step clearly, especially in examinations.

Sign Errors During Simplification

Negative signs frequently appear in reduction formulas.

Always simplify carefully to avoid sign mistakes.

Forgetting the Constant of Integration

Since reduction formulas are used for indefinite integrals, always include:

$+C$

in the final answer.

Stopping Too Early

Continue applying the reduction formula until the remaining integral becomes elementary.

For example:

Do not stop at:

$\int \sin^2 xdx$

if the problem requires complete evaluation.

Ignoring Formula Conditions

Some reduction formulas work only under specific conditions.

Always check:

• Domain restrictions
• Convergence requirements
• Validity of substitutions

Exam Tip

Before applying a reduction formula, ask yourself:

1. What type of integral is this?
2. Which reduction formula applies?
3. What is the value of $n$?
4. Does the formula reduce the power correctly?

Answering these questions first can prevent most common mistakes and make solving reduction-formula problems much easier.

Best Books for Reduction Formulas and Integral Calculus

A strong understanding of integration techniques and reduction formulas is essential for mastering calculus and solving higher-level mathematics problems.

Shortcut Tips and Tricks for Integration by Parts

These shortcut techniques can simplify lengthy integration problems and help identify the most efficient solution method.

Important Formula Table

This formula table contains some of the most frequently used reduction formulas in integration by parts and integral calculus.

Common Integration by Parts Formula

These tables serve as quick revision resources for area formulas, area of parallelogram, reduction formulas, integration by parts, mensuration, geometry, and calculus preparation.

Solved Examples Based on Integration by Reduction Formula

Example 1: Evaluate $\int \sin ^4 x d x$

1) $\frac{1}{8}+\frac{\sin 4 x}{32}+\frac{\sin 2 x}{2}+C$
2) $\frac{\sin (4 x)-8 \sin (2 x)+12 x}{32}+C$
3) $\frac{1}{8}-\frac{\sin 4 x}{32}+\frac{\sin 2 x}{2}+C$

4) none of these

Solution
$
\begin{aligned}
& I=\int \sin ^4 x d x \\
& \Rightarrow I=\int\left(\frac{1-\cos 2 x}{2}\right)^2 d x \\
& =\frac{1}{4} \int\left(1-2 \cos 2 x+\cos ^2 2 x\right) d x \\
& =\frac{1}{4} x-\frac{1}{2} \int \cos 2 x d x+\frac{1}{4} \int \cos ^2 2 x d x \\
& =\frac{1}{4} x-\frac{1}{4} \sin 2 x+\frac{1}{8}\left(x+\frac{1}{4} \sin (4 x)\right) \\
& =\frac{\sin (4 x)-8 \sin (2 x)+12 x}{32}+C
\end{aligned}
$

Hence, the answer is the option (2).

Example 2: Find the value of $\int \sin ^2 x \cos ^2 x d x$

1) $\frac{x}{2}-\frac{\sin 2 x}{2}+C$
2) $\frac{x}{4}-\frac{\sin 4 x}{4}+C$
3) $\frac{1}{8}\left(x-\frac{1}{4} \sin (4 x)\right)+C$
4) $\frac{x}{2}-\frac{\sin 2 x}{4}+C$

Solution

$=\int \sin ^2 x \cos ^2 x d x=\int \frac{\sin ^2(2 x)}{4} d x=\int \frac{1-\cos (4 x)}{8} d x=\frac{1}{8}\left(x-\frac{1}{4} \sin (4 x)\right)+C$

Hence, the answer is the option (3).

Example 3: $\int\left(\tan ^2 x+\cot ^2 x\right) d x$

1) $\tan x+\cot x+C$

2) $\tan x+\cot x-2+C$

3) $-2 x+\tan (x)-\cot (x)+C$

4) $\sec x+\csc x+C$

Solution

$\begin{aligned} & I=\int \tan ^2(x) d x+\int \cot ^2(x) d x \\ & \int \tan ^2(x) d x=\int-1+\sec ^2(x) d x=-x+\tan (x) \\ & \int \cot ^2(x) d x=\int-1+\csc ^2(x) d x=-x-\cot (x) \\ & I=-2 x+\tan (x)-\cot (x)+C\end{aligned}$

Hence, the answer is the option (3).

Example 4: Integrate $\int \tan ^4 x d x$

1) $\frac{\tan ^3 x}{3}-\tan x+C$
2) $\frac{\tan ^3 x}{3}-\tan x+x+C$
3) $\frac{\tan ^3 x}{3}+C$
4) $\frac{\tan ^3 x}{3}+\tan x+C$

Solution

$\begin{aligned} & \int \tan ^4 x d x \\ & =\int \tan ^2 x\left(\sec ^2 x-1\right) d x \\ & =\int \tan ^2 x \sec ^2 x d x-\int \tan ^2 x d x \\ & =\int \tan ^2 x \sec ^2 x d x-\int\left(\sec ^2 x-1\right) d x \\ & =\frac{\tan ^3 x}{3}-\tan x+x+C\end{aligned}$

Hence, the answer is the option (2).

Example 5: Integrate $\int \cot ^5 x d x$

1) $\frac{\cot ^4 x}{4}-\frac{\cot ^2 x}{2}+C$

2) $-\ln (\csc (x))-\frac{\csc ^4(x)}{4}+\csc ^2(x)+c$

3) $\frac{\cot ^4 x}{4}-\cot ^2 x+\ln \sin x+C$

4)None of these

Solution

$\begin{aligned} & \int \cot ^5 x d x=\int \cot x\left(\csc ^2 x-1\right)^2 d x \\ & u=\csc (x): \mathrm{d} x=-\frac{1}{\cot (x) \csc (x)} \mathrm{d} u \\ & =-\int \frac{\left(u^2-1\right)^2}{u} \mathrm{~d} u \\ & =-\ln (\csc (x))-\frac{\csc ^4(x)}{4}+\csc ^2(x)+c\end{aligned}$

Hence, the answer is the option (2).

List of Topics Related to the Reduction Formulas For Integration by Parts

The section explores important topics that build on reduction formulas used in integration by parts. It offers insights into techniques for simplifying irrational algebraic functions, connecting foundational ideas like indefinite and definite integrals with their practical applications.

Application of Integrals

Integral of Particular Functions

Indefinite Integrals

Integration by parts

Application of Inequality in Definite Integration

NCERT Resources

This section compiles all vital NCERT resources for Chapter 7 - Integrals, offering a one-stop guide for effective learning and exam readiness. From detailed notes and step-by-step solutions to challenging exemplar problems, these materials help build a strong foundation in key integration techniques and boost confidence for both board and competitive exams.

NCERT Class 12 Maths Notes for Chapter 7 - Integrals

NCERT Class 12 Maths Solutions for Chapter 7 - Integrals

NCERT Class 12 Maths Exemplar Solutions for Chapter 7 - Integrals

Practice Questions based on Reduction Formulas For Integration by Parts

This section offers a rich set of custom practice questions exploring reduction formulas and integration by parts. Through hands-on exercises involving different algebraic and irrational integrals, you can steadily enhance your technical skills, deepen your understanding, and gain the confidence needed to tackle challenging integration problems independently.

Reduction Formula- Practice Question MCQ

We have provided below the practice questions related to different concepts of integration to improve your understanding: