91Ó°ÊÓ

The given expression is

$$\begin{gathered} \left(a\right){∫}_0^{\mathrm{π}}\sin x\mathrm{δ}\left(x-\frac{\mathrm{π}}{2}\right)dx \\ \left(b\right){∫}_0^{\mathrm{π}}\sin x\mathrm{δ}\left(x+\frac{\mathrm{π}}{2}\right)dx \\ \left(c\right){∫}_{-1}^1{e}^{3x}{\mathrm{δ}}^{\text{'}}\left(x\right)dx \\ \left(d\right){∫}_0^{\mathrm{π}}\cosh x{d}^{\text{'}\text{'}}(x-1)dx \end{gathered}$$

Integration by partsor partial integration is a process that finds theintegralof aproductoffunctionsin terms of the integral of the product of theirderivativeandantiderivative.

The calculation is

${∫}_0^{\mathrm{π}}\sin x\mathrm{δ}\left(x-\frac{\mathrm{π}}{2}\right)dx=(\sin x{)}_{s-x_3}$

Here $x_0=\frac{\mathrm{Ï€}}{2},a=0,b=\mathrm{Ï€}$and$0<\frac{\mathrm{Ï€}}{2}<\mathrm{Ï€}$

$$\begin{gathered} =(\sin x{)}_{x-\frac{\mathrm{Ï€}}{2}} \\ =\sin \frac{\mathrm{Ï€}}{2} \\ =1 \end{gathered}$$

Thus, ${∫}_0^{\mathrm{π}}\sin x·\mathrm{δ}\left(x-\frac{\mathrm{π}}{2}\right)dx$is 1.

The calculation is

$$\begin{gathered} {∫}_0^{\mathrm{π}}\sin x\mathrm{δ}\left(x+\frac{\mathrm{π}}{2}\right)dx \\ {∫}_0^{\mathrm{π}}\sin x\mathrm{δ}\left(x-\left(-\frac{\mathrm{π}}{2}\right)\right)dx \end{gathered}$$

Here $x_0=\frac{-\mathrm{π}}{2},a=0,b=\mathrm{π}$and$\frac{-\mathrm{π}}{2}∉(0,\mathrm{π})$

$=0$

Thus, ${∫}_0^{\mathrm{π}}\sin x·\mathrm{δ}\left(x+\frac{\mathrm{π}}{2}\right)dx$is 0.

The calculation is

${∫}_{-1}^1{e}^{3x}\mathrm{δ}\text{'}\left(x\right)dx={∫}_{-1}^1{e}^{3x}·\mathrm{δ}\text{'}(x-0)dx$

Where $\mathrm{Ï•}\left(x\right)=e^{\text{3x}}$and $a=0$

$$\begin{gathered} =-\mathrm{Ï•}\text{'}\left(x\right) \\ =-{\left(e^{3x}\right)}^{\text{'}} \\ =-{\left[3e^{3x}\right]}_{x-0} \\ =-3e^0 \\ =-3 \end{gathered}$$

Thus, ${∫}_{-1}^1{e}^{3x}{e}^{\text{'}}\left(x\right)dx$is -3.

The calculation is

${∫}_0^x\cosh x·\mathrm{δ}\text{'}\text{'}(x-1)dx$

Where$\mathrm{Ï•}\left(x\right)=\cosh x$and$a=1$

$$\begin{gathered} \mathrm{Ï•}\text{'}\left(x\right)=\left(\cosh x\right)\text{'} \\ =\left(\sinh x\right) \\ \mathrm{Ï•}\text{'}\text{'}\left(x\right)=\left(\sinh x\right)\text{'} \\ =(\cosh x{)}_{x-1} \\ \mathrm{Ï•}\text{'}\text{'}\left(x\right)=\cosh 1 \end{gathered}$$

Thus, ${∫}_0^x\cosh x-{\mathrm{δ}}^{\text{'}\text{'}}(x-1)dx$is cosh1.