Get Answers to all your Questions

header-bg qa

Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 60 Maths Textbook Solution.

Answers (1)


I=\frac{4}{5} x+\frac{3}{5} \log (2 \sin x+\cos x)+c


\int \frac{\sin x+2 \cos x}{2 \sin x+\cos x} d x


To solve this statement we have to use formula of sin θ+ 2cos θ  and also substitute method.


\sin x+2 \cos x=A \frac{d}{d x}(\mathrm{D})+\mathrm{B}(\mathrm{D})

   \sin x+2 \cos x=A(2 \cos x-\sin x)+B(2 \sin x+\cos x)

\sin x=>1=2 B-A \ldots . .(1) \quad \cos x=>2=2 A+B \ldots \ldots .(2)

\text { On solving (1) \& (2), }                        A=\frac{3}{5} \quad \& B=\frac{4}{5}

I=\frac{3}{5} \int \frac{2 \cos x-\sin x}{2 \sin x+\cos x}+\frac{4}{5} \int \frac{2 \sin x+\cos x}{2 \sin x+\cos x} d x

 I=\frac{3}{5} \int \frac{d t}{t}+\frac{4}{5} \int d x

I=\frac{3}{5} \log t+\frac{4}{5} x+c

I=\frac{4}{5} x+\frac{3}{5} \log (2 \sin x+\cos x)+c


Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support