Question

Solve each of the integrals in Exercises $21-66$. Some of the integrals require the methods presented in this section, and some do not. (The last four exercises involve hyperbolic functions).

role="math" localid="1649522213988" $\int cscxco{t}^{2}xdx$.

Short answer

The value of the integral $\int cscxco{t}^{2}xdx$is,$-\frac{1}{2}cscx.cotx+\frac{1}{2}\ln \left|cscx+cotx\right|+C$.

Step-by-step solution

Step 1. Given information

$\int cscxco{t}^{2}xdx$.

Step 2. The integration is solved below.

In$\int cscxco{t}^{2}xdx$, first let us try to find the value for $cscx.co{t}^{2}x$.

$$\begin{gathered} cscx.co{t}^{2}x=cscx\left(cs{c}^{2}x-1\right) \\ =cs{c}^{3}x-cscx \end{gathered}$$

Now find the integral value for $cs{c}^{3}x$.

Let $u=cscx,du=-cotxcscx.dx$

$v=-cotx,dv=cs{c}^{2}x$.

Now apply the integrating by parts rule.

$$\begin{gathered} \int cs{c}^{3}x.dx=-cscx.cotx-\int co{t}^{2}x.cscx.dx \\ =-cscx.cotx-\int cscx.\left(cs{c}^{2}x-1\right)dx \\ =-cscx.cotx-\int (cs{c}^{3}x-cscx)dx \\ =-cscx.cotx-\int cs{c}^{3}x.dx+\int cscx.dx \\ 2\int cs{c}^{3}x.dx=-cscx.cotx+\int cscx.dx \end{gathered}$$

Step 3. Now apply the theorem 5.17 in the obtained equation.

$$\begin{gathered} 2\int cs{c}^{3}x.dx=-cscx.cotx-\ln \left|cscx+cotx\right|+C \\ \int cs{c}^{3}x.dx=-\frac{1}{2}cscx.cotx-\frac{1}{2}\ln \left|cscx+cotx\right|+C \end{gathered}$$

Now substitute the known values and also apply theorem$5.17$ in $\int \left(cs{c}^{3}x-cscx\right)dx$.

role="math" localid="1649527490064" $$\begin{gathered} \int \left(cs{c}^{3}x-cscx\right)dx=-\frac{1}{2}cscx.cotx-\frac{1}{2}\ln \left|cscx+cotx\right|-\left(-\ln \left|cscx+cotx\right|\right)+C \\ =-\frac{1}{2}cscx.cotx+\frac{1}{2}\ln \left|cscx+cotx\right|+C \end{gathered}$$