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Home  >>  CBSE XII  >>  Math  >>  Integrals
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$\Large \int\limits_\frac{-\pi}{4}^\frac{\pi}{4}\frac{dx}{1+\cos2x}$ is equal to $(A)\;1\quad(B)\;2\quad(C)\;3\quad(D)\;4$

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Toolbox:
  • If $f(-x)=-f(x)$,then $f(x)$ is an odd function.
  • If $f(-x)=f(x)$,then $f(x)$ is an even function.
  • $1+\cos 2x=2\cos^2x$
  • $\int \sec^2xdx=\tan x+c$
  • $\int_{-a}^a f(x)dx=2\int_0^af(x)dx$ if $f(x)$ is an even function.
Step 1:
Let $I=\int_{\large\frac{-\pi}{4}}^{\large\frac{\pi}{4}}\large\frac{dx}{1+\cos 2x}$
If $f(x)$ is replaced by $f(-x)$ and if $f(-x)=f(x)$,then the function is said to be an even function.
Consider $1+\cos 2x$
Let $f(x)=1+\cos 2x$
$f(-x)=1+\cos( -2x)$
But $\cos(-x)=\cos x$
Hence $1+\cos(-2x)=1+\cos 2x$
Hence it is an even function.
Step 2:
$I=\int_{-a}^a f(x)dx=2\int_0^af(x)dx$ if $f(x)$ is an even function.
Applying this property,
$I=2\int_0^{\Large\frac{\pi}{4}}\large\frac{dx}{1+\cos 2x}$
$1+\cos2x=2\cos^2x$
$I=2\int_0^{\Large\frac{\pi}{4}}\large\frac{dx}{\cos^2x}=\int_0^{\Large\frac{\pi}{4}}$$\sec^2xdx.$
Step 3:
On integrating we get,
$I=[\tan x]_0^{\Large\frac{\pi}{4}}$
$=\tan\large\frac{\pi}{4}\normalsize-\tan 0$
But $\tan\large\frac{\pi}{4}=1$ and $\tan 0=0.$
Hence I=1.
Hence the correct answer is (A).
answered Apr 24, 2013 by sreemathi.v
 

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