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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Integrate : $\int \large\frac{\sqrt {\cos^{-1}(4x)}}{\sqrt {1-16x^2}}$$dx$

$(a)\;-\frac{1}{6} \sqrt {\frac{\pi}{2} - \sin ^{-1} (4x)}+\cos ^{-1}x+c \qquad(b)\;-\frac{1}{6} \sqrt {\frac{\pi}{2} - \sin ^{-1} (4x)}+c \qquad(c)\;\sqrt { \sin ^{-1} (4x)} \qquad (d)\;None$
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1 Answer

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$\cos ^{-1} (4x)=\large\frac{z}{2}$$ - \sin ^{-1} (4x)$
$\large\frac{z}{2}-$$ \sin ^{-1} (4x)=t$
differentiate with $(x) => 0-\large\frac{1}{\sqrt {1- 16x^2}}$$ \times 4dx=dt$
=> $ \large\frac{dx}{\sqrt {1-16x^2}}=-\large\frac{dt}{4}$
=> $\int \sqrt {t} -\large\frac{dt}{4}$
=> $ -\large\frac{2}{3} \times \frac{t^{3/2}}{4}$$+c$
$-\large\frac{1}{6} \sqrt {\frac{\pi}{2} - \normalsize \sin ^{-1} (4x)}+c$
Hence b is the correct answer.
answered Dec 23, 2013 by meena.p
 
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