If $\int \frac{\sin x}{3+4 \cos ^2 x} \mathrm{~d} x=\mathrm{A} an ^{-1}(\mathrm{~B} \cos x)+\mathrm{C}$, (where $\mathrm{c}$ is a constant of integration), then the value of $\mathrm{A}+\mathrm{B}$ is

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If $\int \frac{\sin x}{3+4 \cos ^2 x} \mathrm{~d} x=\mathrm{A} 	an ^{-1}(\mathrm{~B} \cos x)+\mathrm{C}$, (where $\mathrm{c}$ is a constant of integration), then the value of $\mathrm{A}+\mathrm{B}$ is

Aicharya · Accepted Answer

Let $\mathrm{I}=\int \frac{\sin x}{3+4 \cos ^2 x} \mathrm{~d} x$ Put $\cos x=\mathrm{t}$ $-\sin x \mathrm{~d} x=\mathrm{dt}$ $	herefore \quad \sin x \mathrm{~d} x=-\mathrm{dt}$ $\begin{aligned} 	herefore \quad \mathrm{I} & =\int \frac{-\mathrm{dt}}{3+4 \mathrm{t}^2} \ & =-1 \int \frac{1}{4 \mathrm{t}^2+3} \end{aligned}$ $=-1 \int \frac{1}{(2 t)^2+(\sqrt{3})^2}$ $=-1 	imes \frac{1}{2 	imes \sqrt{3}} 	an ^{-1}\left(\frac{2 \mathrm{t}}{\sqrt{3}}ight)+\mathrm{c}$ $\mathrm{I}=\frac{-1}{2 \sqrt{3}} 	an ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}ight)+\mathrm{c}$ But $\int \frac{\sin x}{3+4 \cos ^2 x} \mathrm{~d} x=\mathrm{A} 	an ^{-1}(\mathrm{~B} \cos x)+\mathrm{c}$ Comparing above equations, we get $\begin{aligned} \mathrm{A} & =\frac{-1}{2 \sqrt{3}}, \mathrm{~B}=\frac{2}{\sqrt{3}} \ 	herefore \quad \mathrm{A} & +\mathrm{B}=\frac{-1}{2 \sqrt{3}}+\frac{2}{\sqrt{3}}=\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2} \end{aligned}$

If $\int \frac{\sin x}{3+4 \cos ^2 x} \mathrm{~d} x=\mathrm{A} \tan ^{-1}(\mathrm{~B} \cos x)+\mathrm{C}$,…

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