Question

integration of sin(logx)+cos(logx)

Answer 1

you can see your answer

Answer 2

Answer:

xsin(logx)+C

Step-by-step explanation:

We have to find the integration of sin(logx)+cos(logx), therefore

\int sin(logx)dx+\int cos(logx)dx

let \int sin(logx)dx=I_{1} and \int cos(logx)dx=I_{2}, then

I_{1}=\int sin(logx).1dx,

Integrating by parts, wehave

I_{1}=sin(logx).x-\int cos(logx){\times}\frac{1}{x}{\times}xdx

     =xsin(logx)-\int cos(logx)dx

Also, ^{I_{2}}=\int cos(logx)dx

$I_{1}+I_{2}$=xsin(logx)-\int cos(logx)dx+\int cos(logx)dx

                              =xsin(logx)+C