Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need solution for RD Sharma Maths Class 12 Chapter 18 Indefiite Integrals Excercise Fill in the Blanks Question 9

Need solution for RD Sharma Maths Class 12 Chapter 18 Indefiite Integrals Excercise Fill in the Blanks Question 9

Answers (1)

Answer:

-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+c

Hint:

Simplify the function and then integrate it by using formula.

Solution:

I=\int \frac{\sin x}{3+4 \cos ^{2} x} d x

Let \cos x=y

\begin{aligned} &\Rightarrow-\sin x d x=d y \\ &\Rightarrow I=-\int \frac{d y}{3+4 y^{2}} \end{aligned}

\begin{aligned} &\Rightarrow I=-\frac{1}{4} \int \frac{d y}{y^{2}+\frac{3}{4}} \\\\ & \end{aligned}

\Rightarrow I=-\frac{1}{4} \int \frac{d y}{y^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}
\begin{aligned} &I=-\frac{1}{4}\left(\frac{2}{\sqrt{3}}\right) \tan ^{-1}\left(\frac{y}{\frac{\sqrt{3}}{2}}\right)+c \\ &I=-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+c \end{aligned}
 

Posted by

infoexpert27

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

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question