_ ^ 1 a ( a^x a^x)dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{1}{{\log a}}({a^x}\cos {a^x})dx = } $

A
$\sin {a^x} + c$
B
${a^x}\sin {a^x} + c$
✓
$\frac{1}{{{{(\log a)}^2}}}\sin {a^x} + c$
D
$\log \sin {a^x} + c$

✓

Answer

Correct option: C.

$\frac{1}{{{{(\log a)}^2}}}\sin {a^x} + c$

c
(c)$\int_{}^{} {\frac{1}{{\log a}}({a^x}\cos {a^x})\,dx} $
Put ${a^x} = t \Rightarrow {a^x}dx = \frac{{dt}}{{\log a}},$ then it reduces to
$\int_{}^{} {\frac{1}{{{{(\log a)}^2}}}\cos t\,dt} = \frac{1}{{{{(\log a)}^2}}}\sin t + c = \frac{1}{{{{(\log a)}^2}}}\sin {a^x} + c$.

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