Answer
(a) By using the method of differentiating right side of the equation , we conclude that $\int cotxdx=ln|sinx|+C$.
(b) By using the method of substitution, we prove that
$\int cotxdx=ln|sinx|+C$.
Work Step by Step
Show that $\int cotxdx=ln|sinx|+C$
(a) By using the method of differentiating right side of the equation as follows:
Since, $\frac{d}{dx}[A(x)+c]=B(x)$
Then $\int B(x) =A(x)+C$
This implies
$[\frac{d}{dx}ln|sinx|+C]=\frac{d}{dx}ln|sinx|+\frac{d}{dx}C$
$=\frac{1}{sinx}.(cosx)+0$
$=cotx$
Therefore, by using the method of differentiating right side of the equation , we conclude that $\int cotxdx=ln|sinx|+C$.
(b) By using the method of substitution.
For this, first we write $cotx$ in terms of sine and cosine such as:
$\int cotx dx=\int\frac{cosx}{sinx}dx$
This suggests that we should substitute $u=sinx$, since then $du=cosxdx$
Thus, $\int cotx dx=\int\frac{cosx}{sinx}dx=\int\frac{1}{u}du$
This implies
$ln|u|+C=ln|sinx|c$
This result can also be written as
$\int cotxdx=ln|sinx|+C$
Hence, the result is proved.
