integration of sin(logx)+cos(logx)dx
Answer (1)
9th Jan, 2020
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.
Comments (0)
Related Questions
integration zero se Pi lok 1 + cos x dx
117 Views
dy/dx=sin ( x + y) + cos ( x + y )
95 Views
