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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Integral Calculus

Integrate : $\int 5x^2 \tan ^2 (6x^3) dx$

$(a)\;cos (6x^3)+c \qquad(b)\;\sin (6x^3)+c \qquad(c)\;\frac{5}{18} [\cot (6x^3)-(6x^3)]+c \qquad (d)\;\frac{5}{18} [\tan (6x^3)-(6x^3)]+c$

1 Answer

$6x^3=t$=> differentiate with respect to x
$18 x ^2 dx =dt$
=> $\int \large\frac{5}{18}$$ \tan ^2 t dt$
=> $\int \large\frac{5}{18} $$(\sec ^2 t -1)$$dt$
=> $ \large\frac{5}{18} $$(\sec ^2 t -1)$$dt$
=> $\large\frac{5}{18} $$(\tan t -t)+c$
$ \large\frac{5}{18}$$ [\tan (6x^3)-(6x^3)]+c$
Hence d is the correct answer.
answered Dec 24, 2013 by meena.p
edited Mar 13, 2014 by balaji.thirumalai
 
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