Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please Solve RD Sharma Class 12 Chapter 18.17 Indefinite Integrals Exercise Very Short Answer Question 2 Maths Textbook Solution

Please Solve RD Sharma Class 12 Chapter 18.17 Indefinite Integrals Exercise Very Short Answer Question 2 Maths Textbook Solution

Answers (2)

Answer:-

Hint:-To solve this problem, use special integration formula

Given:-  

 Solution:-  Let  

   

Put

                                                

Posted by

infoexpert27

View full answer

Crack CUET with india's "Best Teachers"

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

Answer:- \sin^{-1}\left ( \frac{2x-3}{\sqrt{41}} \right )+c

Hint:-To solve this problem, use special integration formula

Given:- \int \frac{1}{\sqrt{8+3x-x^{2}}}dx 

 Solution:-Let I=\int \frac{1}{\sqrt{8+3x-x^{2}}}dx=\int \frac{1}{\sqrt{-\left ( x^{2}-3x-8 \right )}}dx 

\begin{aligned} &\Rightarrow I=\int \frac{1}{\sqrt{-\left\{x^{2}-2 \cdot x \cdot \frac{3}{2}+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}-8\right\}}} d x \\ &=\int \frac{1}{\sqrt{-\left\{\left(x-\frac{3}{2}\right)^{2}-\frac{9}{4}-8\right\}}} d x=\int \frac{1}{\sqrt{-\left\{\left(x-\frac{3}{2}\right)^{2}-\left(\frac{9+32}{4}\right)\right\}}} d x \end{aligned}   

 

\begin{aligned} &=\int \frac{1}{\sqrt{-\left\{\left(x-\frac{3}{2}\right)^{2}-\frac{41}{4}\right\}}} d x=\int \frac{1}{\sqrt{\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}}} d x \\ &=\int \frac{1}{\sqrt{\left(\frac{\sqrt{41}}{2}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}}} d x \end{aligned}

Put x-\frac{3}{2}=t\Rightarrow dx=dt\: then

I=\int \frac{1}{\sqrt{\left ( \frac{\sqrt{41}}{2} \right )^{2}-t^{2}}}dt=\sin^{-1}\left (\frac{t}{\frac{\sqrt{41}}{2}} \right )+c                                                 \left [ \because \int \frac{1}{\sqrt{a^{2}-x^{2}}}dx=\sin^{-1}\left ( \frac{x}{a} \right )+c \right ]

=\sin^{-1}\left ( \frac{x-\frac{3}{2}}{\frac{\sqrt{41}}{2}} \right )+c=\sin^{-1}\left ( \frac{\frac{2x-3}{2}}{\frac{\sqrt{41}}{2}} \right )+c                                                     \left [ \because t=x-\frac{3}{2} \right ]

Posted by

infoexpert27

View full answer

Similar Questions

Trending Articles/News

anam-khan

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question