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Reduction Formula - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

Reduction Formula (Part 1) is considered one of the most asked concept.
20 Questions around this concept.

Solve by difficulty

DIFFICULT

Evaluate the integral of $x^{5}log^{2}x$

$\frac{x^{6}log^{2}x}{6}+\frac{x^{6}log\ x}{18}+\frac{x^{6}}{108}+C$

$\frac{x^{6}log^{2}x}{6}-\frac{x^{6}log\ x}{18}+\frac{x^{6}}{108}+C$

$\frac{x^{6}log^{2}x}{6}-\frac{x^{6}log\ x}{18}-\frac{x^{6}}{108}+C$

$-\frac{x^{6}log^{2}x}{6}-\frac{x^{6}log\ x}{18}+\frac{x^{6}}{108}+C$

View Solution

Concepts Covered - 2

Reduction Formula (Part 1)

$\mathbf{(a)\;\;\;{\int\;\sin^n x\;dx}}$

$\\\text {Let } \;\;\;\;\;\;\;I_{n}=\int \sin ^{n} x d x=\int \sin ^{n-1} x \sin x d x\\\mathrm{take\;\sin^{n-1}x\;as\;first\;function\;and\;\sin x\;as\;second\;function}\\ \mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}{=-\sin ^{n-1} x \cos x+\int(n-1) \sin ^{n-2} x \cos ^{2} x d x} \\ \mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} {=-\sin ^{n-1} x \cos x+(n-1) \int \sin ^{n-2} x\left(1-\sin ^{2} x\right) d x} \\ \mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} {=-\sin ^{n-1} x \cos x+(n-1) \int\left(\sin ^{n-2} x-\sin ^{n} x\right) d x}\\ \mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=-\sin ^{n-1} x \cos x+(n-1) I_{n-2}-(n-1) I_{n}\\\\ {\therefore \;\;\;\;\;\;\;n I_{n}=-\sin ^{n-1} x \cos x+(n-1) I_{n-2}} \\ {\Rightarrow\;\;\;\;\;\;\;\; I_{n}=-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} I_{n-2}}\\\\ \text { Thus, } \int \sin ^{n} x d x=\frac{-\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} \int \sin ^{n-2} x d x$

$\mathbf{(b)\;\;\;{\int\;\tan^n x\;dx}}$

$\\\text { Let }\;\;\;\;\;\; I_{n}=\int \tan ^{n} x d x\\ \Rightarrow \;\;\;\;\;\;\;\;I_{n} =\int \tan ^{n-2} x \tan ^{2} x d x=\int \tan ^{n-2} x\left(\sec ^{2} x-1\right) d x \\ \mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\int \tan ^{n-2} x \sec ^{2} x-I_{n-2}=\int t^{n-2} d t-I_{n-2} \\\text { where, } \tan x=t\;\; \Rightarrow \;\;\sec ^{2} x \;d x=d t$

$\\ \mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;}I_{n}=\frac{t^{n-1}}{n-1}-I_{n-2}\\ \mathrm{\therefore \;\;\;\;\;\;\;\;\;\;}I_{n}=\frac{\tan ^{n-1} x}{n-1}-I_{n-2}\\\\\Rightarrow \int \tan ^{n} x d x=\frac{\tan ^{n-1} x}{n-1}-\int \tan ^{n-2} x d x$

$\mathbf{(c)\;\;\;{\int\;\cos^n x\;dx}}$

$\\\text {Let } I_{n}=\int \cos ^{n} x d x=\int \cos ^{n-1} x \cos x d x\\\text{Take\;}\cos^{n-1}x \text{ as first function and }\cos x\text{ as second function.}\\\\ \mathrm{\;\;\;\;\;}\;\;\;\; \begin{array}{l}{=\cos ^{n-1} x \sin x+\int(n-1) \cos ^{n-2} x \sin ^{2} x d x} \\ {=\cos ^{n-1} x \sin x+(n-1) \int \cos ^{n-2} x\left(1-\cos ^{2} x\right) d x} \\ {=\cos ^{n-1} x \sin x+(n-1) I_{n-2}-(n-1) I_{n}}\end{array}\\\\\therefore n I_{n}=\cos ^{n-1} x \sin x+(n-1) I_{n-2}\\\\\text { or } \int \cos ^{n} x d x=\frac{\cos ^{n-1} x \sin x}{n}+\frac{n-1}{n} \int \cos ^{n-2} x d x$

$\mathbf{(d)\;\;\;{\int\;\cot^n x\;dx}}$

$\\\text { Let } I_{n}=\int \cot ^{n} x d x=\int \cot ^{n-2} x \cot ^{2} x d x\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;}=\int \cot ^{n-2} x\left(\csc ^{2} x-1\right) d x\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;}=\int \cot ^{n-2} x \csc ^2xd x-I_{n-2} \\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;}=\int t^{n-2} d t-I_{n-2}\\\text { where, } \cot x=t \Rightarrow \csc ^{2} x d x=-d t\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}I_{n}=-\frac{\cot ^{n-1} x}{n-1}-I_{n-2}\\\\\therefore \quad \int \cot ^{n} x d x=-\frac{\cot ^{n-1} x}{n-1}-\int \cot ^{n-2} x d x$

$\mathbf{(e)\;\;\;{\int\;\sec^n x\;dx}}$

$\\\text {Let } I_{n}=\int \sec ^{n} x d x=\int \sec ^{n-2} x \sec ^{2} x d x\\\\\mathrm{Take\;\sec^{n-2}x\;as\;first\;function\;and\;\sec^2x\;as\;a\;second\;function}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;}=\sec ^{n-2} x \tan x-\int(n-2) \sec ^{n-3}x\sec x\tan x\cdot\tan xdx\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;}=\sec ^{n-2} x \tan x-(n-2) \int \sec ^{n-2} x\left(\sec ^{2} x-1\right) d x\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;}=\sec ^{n-2} x \tan x-(n-2) I_{n}+(n-2) I_{n-2}\\\\\Rightarrow (n-1) I_{n}=\sec ^{n-2} x \tan x+(n-2) I_{n-2}\\\\\mathrm{or\;\;\;\;\;\;\;\;\;}I_{n}=\frac{\sec ^{n-2} x \tan x}{(n-1)}+\frac{(n-2)}{(n-1)} I_{n-2}\\\\\therefore \int \sec ^{n} x d x=\frac{\sec ^{n-2} x \tan x}{(n-1)}+\frac{(n-2)}{(n-1)} \int \sec ^{n-2} x d x$

$\mathbf{(f)\;\;\;{\int\;\csc^n x\;dx}}$

$\begin{array}{l}{\text { Let } I_{n}=\int \csc ^{n} x d x=\int \csc ^{n-2} x \csc ^{2} x d x} \\ {\quad\quad\quad\quad=\csc ^{n-2} x(-\cot x)-\int(n-2) \csc ^{n-2} x\left(\csc ^{2} x-1\right) d x} \\ {\quad\quad\quad\quad=-\csc ^{n-2} x\left(-\cot x)-(n-2) \int\left(\csc ^{n} x-\csc ^{n-2} x\right) d x\right.} \\ {\quad\quad\quad\quad=-\csc ^{n-2} x \cot x-(n-2) I_{n}+(n-2) I_{n-2}} \\\\ {\therefore \quad(n-1) I_{n}=-\csc ^{n-2} x \cot x+(n-2) I_{n-2}} \\\\ {\text { or } \quad I_{n}=-\frac{\csc ^{n-2} x \cot x}{n-1}+\frac{n-2}{n-1} I_{n-2}} \\\\ {\therefore \int \csc ^{n} x d x=-\frac{\csc ^{n-2} x \cot x}{n-1}+\frac{n-2}{n-1} \int \csc ^{n-2} x d x}\end{array}$

Reduction Formula (Part 2)

$\mathbf{Integration\;of\;the\;type\;\;\;\int \cos^m x\; \sin nx\;dx}$

$\\\mathrm{Let\;I_{m,n}=\int\cos^m x\sin nx\;dx}\\\\\text{To\;evaluate\;this\;integral, we will use integration by parts method}\\\mathrm{Here,\;take\;\cos^m x\;as\;first\;function\;and \;\sin nx\;as\;second\;function.}\\\\\mathrm{\;\;\;\;\;\;\;\;\;}=-\frac{\cos ^{m} x \cos n x}{n}-\frac{m}{n} \int \cos ^{m-1} x \sin x \cos n x d x\\\mathrm{\;\;\;\;\;\;\;\;\;}=-\frac{\cos ^{m} x \cos n x}{n}-\frac{m}{n} \int \cos ^{m-1} x\{\sin n x \cos x-\sin (n-1) x\} d x\\ \text { [using } \sin (n-1) x=\sin n x \cos x-\cos n x \sin x\\\mathrm{\;\;\;\;\;\;}\Rightarrow \sin x \cos n x=\sin n x \cos x-\sin (n-1) x]\\\\\mathrm{\;\;\;\;\;\;\;\;\;}=-\frac{\cos ^{m} x \cos n x}{n}-\frac{m}{n} \int \cos ^{m} x \sin n x d x+\frac{m}{n} \int \cos ^{m-1} x \sin (n-1) x d x\\\mathrm{\;\;\;}I_{m, n}=-\frac{\cos ^{m} x \cos n x}{n}-\frac{m}{n} I_{m, n}+\frac{m}{n} I_{m-1, n-1}$

$\\ {\Rightarrow \frac{m+n}{n} I_{m, n}=-\frac{\cos ^{m} x \cos n x}{n}+\frac{m}{n} I_{m-1, n-1}} \\ {\text { or }\;\;\;\;\;\; \quad I_{m, n}=-\frac{\cos ^{m} x \cos n x}{m+n}+\frac{m}{m+n} I_{m-1, n-1}}$

NOTE :

In the simmilar way we can also prove the following result

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