Question

$\int \frac{1}{\log a}\left(a^x\cos a^x\right)dx=$

• A. $\frac{1}{(\log a{)}^2}\sin a^x+C$
• B. $(\log a{)}^2\sin a^x+C$
• C. $\frac{1}{\log a}\sin a^x+C$
• D. $\sin a^x+C$
• E. $a^x\sin a^x+C$

Answer 1

Answer: (A)

Put $a^x=v\Rightarrow a^{x}dx=\frac{dv}{\log a^{{}^{\prime }}}$ then it reduces to
$\int \frac{l}{(\log a{)}^2}\cos vdv=\frac{l}{(\log a{)}^2}\sin v+C$
$=\frac{1}{(\log a{)}^2}\sin a^x+C$