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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Integrate : $\int \sin ^2 x. \cos ^2 x \;dx$

\[\begin {array} {1 1} (a)\;\frac{x}{2}-\frac{\cos 4x}{32}+c \\ (b)\;\frac{x}{8}-\frac{\sin 4x}{32}+c \\ (c)\;\frac{x}{8}+\frac{\cos 4x}{16}+c \\ (d)\;\frac{x}{8}+\frac {\cos 4x}{16}+c \end {array}\]
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1 Answer

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$\large\frac{1}{4}$$ \int 4 \sin ^2 x \cos^2 x dx$
=> $\large\frac{1}{4} $$(2 \sin x \cos x)^2 dx$
=> $\large\frac{1}{4}$$ \int \sin ^2 2x dx$
=> $\large\frac{1}{4} \times \frac{1}{2}$$\bigg\{(1- \cos 4x)dx$
=> $\large\frac{1}{8}$$\bigg\{(1- \cos 4x)dx$
=> $\large\frac{1}{8}$$\bigg\{x-\large\frac{\sin 4x}{4} \bigg\}+c$
Hence b is the correct answer.
answered Dec 18, 2013 by meena.p
 
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