Find value of integral ∫(𝐬𝐢𝐧 𝐱)/(3+4𝐜𝐨𝐬² 𝐱 ) 𝐝𝐱

To find the value of this integral 

Consider the integral $\int \frac{sin x }{3+ 4 cos^2 x}$ we can rewrite it as

\[\int \frac{\textbf{sinx}}{{3+ 4 \cos^2 x}}=\int \frac{\sin x}{3+ (2\cos x)^2}\]

For finding the value put  $ 2\cos x=t$

By differentiating with respect to x then we will get -2 sin x  = dt/dx \[\Rightarrow \sin x dx = -dt/2\]

Hence

\[\int \frac{\textbf{sinx}}{{3+ 4 \cos^2 x}}=\int \frac{-dt}{2(\sqrt{3})^2+ (t)^2)}  \]

We have  the formulae $\int \mathbf{\frac{1}{a^2 + x^2}}dx = \frac{1}{a} \tan ^{-1} \frac{x}{a}+C$ .

Now by using that we can find value of the given integral 

i.e., 

 \[\int \frac{-dt}{2((\sqrt{3})^2 +t^2)}=\frac{-1}{2\sqrt{3}}\tan ^{-1}\frac{t}{\sqrt{3}}+C\] 

Now by substituting value of t we will get the value of integral as

\[\int \frac{\textbf{sinx}}{{3+ 4 \cos^2 x}} dx=\frac{-1}{2\sqrt{3}}\tan ^{-1}(\frac{2 \cos x}{\sqrt{3}})+C\]

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