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  • Need Solution for R.D. Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 19 Maths Textbook Solution.

Need Solution for R.D. Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions  Question 19  Maths Textbook Solution.

Answers (1)

Answer:

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

Given:

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

Hint:

You must know about the derivative of cos x and \int \frac{1}{1+x^{2}} d x

Explanation:

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

             =\int \frac{-d t}{3+4 t^{2}}                                                                [\text { Put } \cos x=t \Rightarrow-\sin x d x=d t \Rightarrow \sin x d x=-d t]

            \begin{aligned} &=\frac{-1}{4} \int \frac{d t}{t^{2}+\frac{3}{4}} \\ =& \frac{-1}{4} \int \frac{d t}{(t)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} \\ =& \frac{-1}{4} \cdot \frac{1}{\frac{\sqrt{3}}{2}} \tan ^{-1}\left(\frac{t}{\frac{\sqrt{3}}{2}}\right)+c \ldots \ldots \int \frac{1}{x^{2}+a^{2}} d x=\tan ^{-1} \frac{x}{a}+c \\ =& \frac{-1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 t}{\sqrt{3}}\right)+C \\ =& \frac{-1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+C \end{aligned}

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