Question

$\int \frac{\log x.sin(1+(\log x{)}^2)}{x}dx=$

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

Answer 1

The correct option is A $-\frac{1}{2}\cos (1+(\log x{)}^2)+c$

$\int \log x\cdot \frac{sin(1+(logx{)}^2)}{x}\cdot dx$

$\log x=t$

$\frac{1}{x}dx=dt$

$\int t\sin (1+t^2)\cdot dt$

$\int tsin(1+t^2)dt$

Let $t^2=q$

$2tdt=dq$

$tdt=\frac{1}{2}dq$

$=\int \frac{1}{2}sin(1+q)$

$=\frac{1}{2}\int sin(1+q)dq$

$=-\frac{1}{2}cos(1+q)+c$

$=-\frac{1}{2}cos(1+t^2)+c$

$\int \frac{logx\cdot sin(1+(logx{)}^2)}{x}=-\frac{1}{2}cos[1+(logx{)}^2]+c$

Hence, option 'A' is correct.