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# $\Large\int\frac{\sin x}{3+4\cos^2x}$$dx=___________. Can you answer this question? ## 1 Answer 0 votes Toolbox: • If f(x) is substituted by f(t) and if f'(x)dx=f'(t)dt • Then \int f(x)dx=\int f(t)dt • \int\large\frac{dx}{x^2+a^2}=\large\frac{1}{a}$$\tan^{-1}\big(\large\frac{x}{a}\big)$$+c. Step 1: Let I=\int\large\frac{\sin x}{3+4\cos^2x}$$dx$
Put $\cos x=t$.On integrating with respect to t we get
$-\sin x dx=dt.$
On substituting this we get
$I=-\int\large\frac{dt}{3+4t^2}$
$3+4t^2$ can be written as $4(\large\frac{3}{4}$$+t^2) I=-\large\frac{1}{4}\int\large\frac{dt}{t^2+(\Large\frac{\sqrt 3}{2})^2} -\large\frac{1}{4}\int\large\frac{dt}{t^2+(\Large\frac{\sqrt 3}{2})^2} is of the form \int\large\frac{dx}{x^2+a^2} \Rightarrow \large\frac{1}{a}\normalsize\tan^{-1}\big(\large\frac{x}{a}\big) Step 2: Hence I=-\large\frac{1}{4}\int\frac{dt}{t^2+(\Large\frac{\sqrt 3}{2})^2} Where x=t and a=\large\frac{\sqrt 3}{2} On integrating we get, I=-\large\frac{1}{4}\begin{bmatrix}\frac{1}{\sqrt 3/2}\tan^{-1}\big(\frac{t}{\sqrt 3/2}\big)\end{bmatrix}=-\large\frac{1}{4}\begin{bmatrix}\frac{2}{\sqrt 3}\normalsize \tan^{-1}\large\frac{2t}{\sqrt 3}\end{bmatrix} Substituting for t we get, I=-\large\frac{1}{4}\begin{bmatrix}\frac{2}{\sqrt 3}\normalsize\tan^{-1}\large\frac{2\cos x}{\sqrt 3}\end{bmatrix}=-\frac{2}{\sqrt 3}$$\tan^{-1}\large\frac{2\cos x}{\sqrt 3}$.