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

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

Answers (1)

Answer:

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

Given:

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

 Hint:

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

Solution: 

\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

 

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