Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 44 maths

explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 44 maths

Answers (1)

Answer: \frac{1}{2} \tan ^{2} x+\frac{1}{4} \tan ^{4} x+c

Hint: Use substitution method to solve this integral.

Given:   \int \sec ^{4} x \tan x\; d x

Solution:

        \begin{aligned} &\text { Let } I=\int \sec ^{4} x \tan x \; d x \\ &\text { Put } \tan x=t \Rightarrow \sec ^{2} x \; d x=d t \\ &\Rightarrow d x=\frac{d t}{\sec ^{2} x} \text { then } \end{aligned}

        I=\int \sec ^{4} x \cdot t \cdot \frac{d t}{\sec ^{2} x}=\int \sec ^{2} x \cdot t\; d t

           =\int\left\{1+\tan ^{2} x\right\} . t \; d t \quad\left[\begin{array}{l} \because \sec ^{2} x-\tan ^{2} x=1 \\ \Rightarrow \sec ^{2} x=1+\tan ^{2} x \end{array}\right]

           \begin{aligned} &=\int\left(1+t^{2}\right) . t\; d t \quad[\because \tan x=t] \\ &=\int\left(t+t^{2} t\right) d t=\int\left(t+t^{3}\right) d t \\ &=\int t \; d t+\int t^{3} d t \end{aligned}

           =\frac{t^{1+1}}{1+1}+\frac{t^{3+1}}{3+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]

           \begin{aligned} &=\frac{t^{2}}{2}+\frac{t^{4}}{4}+c \\ &=\frac{1}{2} \tan ^{2} x+\frac{1}{4} \tan ^{4} x+c \quad[\because t=\tan x] \end{aligned}

Posted by

infoexpert26

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Date Sheet 2027: Class 10, 12 timetable expected soon on cbse.gov.in; check LOC dates, passing marks
anam-khan

CBSE Date Sheet 2027 LIVE: 10th, 12th board exam schedule soon; simple papers, passing marks
anam-khan

CBSE Date Sheet 2027 LIVE: Class 10, 12 board exam schedule soon; passing marks, paper pattern

Latest Question