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  • 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}
 

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