Let I= ∫[sin(log x) + cos(log x)]dx

Put log x = t

⇒x = et

⇒dx = et dt

Now, I = ∫[sin t + cos t].et dt

=∫et . sin t dt + ∫et cos t dt

=sin t × ∫et dt − ∫[ddt(sin t) × ∫et dt]dt + ∫et cos t dt + C

=sin t . et − ∫cost t . et dt + ∫et cos t dt + C

=et . sin t + C

=x . sin(log x) + C

in the above sum et is e power t...kindly notice it.