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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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$\int \limits_0^1 |\sin 2zx| dx$

\[\begin {array} {1 1} (a)\;0 \\ (b)\;\frac{-1}{z} \\ (c)\;\frac{1}{z} \\ (d)\;\frac{2}{z} \end {array}\]
Can you answer this question?
 
 

1 Answer

0 votes
$\sin 2zx=0$
$2zx= n,\pi$
$x= \large\frac{n}{2}, $$ n \in o,1$
$\int \limits_0^{1/2} \sin 2zx dx + \int \limits_{1/2} ^1 -(\sin 2zx)dx$
$-\bigg[ \large\frac{\cos 2zx}{2z} \bigg]_0^{1/2}+ \bigg[ \large\frac{\cos 2 \pi x}{2x}\bigg]^1 _{1/2}$
$ -\large\frac{1}{2z}$$ [ \cos z - \cos 0 - \cos 2z+ \cos z ]$
$-\large\frac{1}{2z} $$[-2-1+ \cos z]$
$\qquad= \large\frac{-1}{z}$
Hence b is the correct answer.
answered Dec 16, 2013 by meena.p
 
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