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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Integrate : $\int \sin 4x \cos 4x\;dx$

\[\begin {array} {1 1} (a)\;\frac{1}{16}\cos 8 x +c \\ (b)\;\frac{1}{16}\sin 2 x +c \\ (c)\;\frac{-1}{16}\cos 8 x +c \\ (d)\;\frac{-1}{16}\sin 2 x +c \end {array}\]
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1 Answer

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$\large\frac{1}{2} $$\int 2 \sin 4x \cos 4x \;dx$
$\{ \sin 2 \theta= 2 \sin \theta \cos \theta\}$
=> $\large\frac{1}{2}$$ \int \sin 8 x \;dx$
=>$ -\large\frac{1}{2} $$\cos 8x \times \large\frac{1}{8} $$+c$
$\large\frac{-1}{16}$$\cos 8 x +c $
Hence c is the correct answer.


answered Dec 19, 2013 by meena.p
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