Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 2 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 2 Maths Textbook Solution.

Answers (1)

Answer:

\frac{1}{2} \log \left(\tan \left(\frac{x}{2}+\frac{\pi}{12}\right)\right)+C

Given:

\int \frac{1}{\cos x+\sqrt{3} \sin x} d x

Hint:

You must know about the \int \cos e c x d x

Explanation:

Let I=\int \frac{1}{2\left(\sin x \cdot \frac{\sqrt{8}}{2}+\cos x \cdot \frac{1}{2}\right)} d x

          =\frac{1}{2} \int \frac{d x}{\sin x \cos _{6}^{\pi}+\cos x \sin \frac{\pi}{6}}                                \left[\because \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2} ; \sin \frac{\pi}{6}=\frac{1}{2}\right]

          =\frac{1}{2} \int \frac{d x}{\sin \left(x+\frac{\pi}{6}\right)}                                                    [\sin (A+B)=\sin A \cos B+\cos A \sin B]

         =\frac{1}{2} \int \operatorname{cosec}\left(x+\frac{\pi}{6}\right) d x                                                   \left[\because \sin x=\frac{1}{\cos \theta c x}\right]

         =\frac{-1}{2} \log \left(\operatorname{cosec}\left(x+\frac{\pi}{6}\right)+\cot \left(x+\frac{\pi}{6}\right)\right)+C        \left[\int \operatorname{cosec}(x) d x=-\log (\operatorname{cosec} x+\cot x)+c\right]

Now,

\begin{aligned} \cos e c x+\cot x &=\frac{1}{\sin x}+\frac{\cos x}{\sin x} \\ &=\frac{1+\cos x}{\sin x} \\ &=\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos _{2}^{x}} \end{aligned}                                                     \left[\because 1+\cos 2 \theta=2 \cos ^{2} \theta ; \sin \theta=2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right]

                              =\cot \frac{x}{2}

\therefore \text { From } E q \cdot(i)

\begin{aligned} &I=\frac{-1}{2} \log \left(\cot \left(\frac{x+\frac{\pi}{6}}{2}\right)\right)+C \\ &=\frac{1}{2} \log \left[\cot \left(\frac{x}{2}+\frac{\pi}{12}\right)\right]^{-1}+C \\ &=\frac{1}{2} \log \left(\tan \left(\frac{x}{2}+\frac{\pi}{12}\right)\right)+C \end{aligned}

Posted by

infoexpert21

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