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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Integrate : $\int \large\frac{\sin 4x}{\sqrt {8+\sin ^4 2x}}$$dx$

$(a)\;\log | \sin^2 2x -\sqrt {\sin ^4 2x+8}|+c \\(b)\; \log | \sin^2 2x +\sqrt {\sin ^4 2x+8}|+c \\(c)\; \frac{1}{2} \log | \sin^2 2x -\sqrt {\sin ^4 2x+8}|+c \\ (d)\;\frac{1}{2}\log | \sin^2 2x +\sqrt {\sin ^4 2x+8}|+c$

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1 Answer

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$\sin ^2 2x =t$
differentiate with x
$2 \sin 2x . \cos 2x .2 dx=dt$
=> $2 \sin 4x dx=dt$
=> $\large\frac{1}{2} \times \int \large\frac{dt}{(8+t^2)}$
=> $ \large\frac{1}{2} $$\times \log |t +\sqrt {t^2+ 8}|+c$
$\large\frac{1}{2}$$\log | \sin^2 2x +\sqrt {\sin ^4 2x+8}|+c$
Hence d is the correct answer.
answered Dec 31, 2013 by meena.p
 
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