Q. 92 - Calculus

Question

Prove the integration formula

$\int \cot x\mathrm{d}x=-\ln \left|\csc x\right|+C$

(a) by using algebra and integration by substitution to find $\int \cot x\mathrm{d}x$;

(b) by differentiating $-\ln \left|\csc x\right|$.

Step-by-Step Solution

Answer

(a) After using algebra and integration by substitution $\int \cot x\mathrm{d}x=-\ln \left(\mathrm{cscx}\right)+\mathrm{C}$.

(b) After differentiating $-\frac{d}{dx}\ln \left|\csc x\right|=\cot x$.

Step 1. Given Information

Prove the integration formula

$\int \cot x\mathrm{d}x=-\ln \left|\csc x\right|+C$

(a) by using algebra and integration by substitution to find $\int \cot x\mathrm{d}x$;

(b) by differentiating $-\ln \left|\csc x\right|$.

Part (a) Step 1. Using algebra and integration by substitution to find ∫ cot x d x

We can write as 

$\int \cot x\mathrm{d}x=\int \frac{\cos x}{\sin x}\mathrm{d}x$

Let

$$\begin{gathered} u=\sin x \\ \frac{du}{dx}=\cos x \\ du=\cos xdx \end{gathered}$$

Part (a) Step 2. Now the integral after substitution.

$$\begin{gathered} \int \cot x\mathrm{d}x=\int \frac{1}{u}\mathrm{du} \\ \int \cot x\mathrm{d}x=\ln \left(\mathrm{u}\right)+\mathrm{C} \\ \int \cot x\mathrm{d}x=\ln \left(\sin x\right)+\mathrm{C} \\ \int \cot x\mathrm{d}x=-\ln \left(\mathrm{cscx}\right)+\mathrm{C} \end{gathered}$$

Part (b) Step 1. By differentiating − ln | csc x |

$$\begin{gathered} -\frac{d}{dx}\ln \left|\csc x\right|=-\frac{d}{dx}\ln \left|\csc x\right|·\frac{d}{dx}\csc x \\ -\frac{d}{dx}\ln \left|\csc x\right|=-\frac{1}{\csc x}·(-\csc x\cot x) \\ -\frac{d}{dx}\ln \left|\csc x\right|=\cot x \end{gathered}$$