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Home  >>  CBSE XII  >>  Math  >>  Integrals
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$\Large\int\frac{\sin x}{3+4\cos^2x}$$dx$=___________.

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  • 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}$$dx$
Put $\cos x=t$.On integrating with respect to t we get
$-\sin x dx=dt.$
On substituting this we get
$3+4t^2$ can be written as $4(\large\frac{3}{4}$$+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}$.
answered Apr 25, 2013 by sreemathi.v
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