find the area y=logx,y=sin^4(x)x=0

Here the point of intersection of the two curve is ( 1.99 , 0.690)
We have to find the area of Shaded region.
So the area of  region is :
$$\begin{gathered} {\int }_1^{1.99}\ln xdx + {\int }_{1.99}^{2.8}{\sin }^{4}xdx \\ = {\mathrm{I}}_1 + {\mathrm{I}}_2 \\ \mathrm{For} {\mathrm{I}}_1 \mathrm{take} \mathrm{lnx} \mathrm{as} \mathrm{first} \mathrm{function} \mathrm{and} 1 \mathrm{as} \mathrm{second} \mathrm{function} \mathrm{and} \mathrm{use} \mathrm{by} \mathrm{part} \\ \mathrm{Hence} {\int }_1^{1.99}\ln \mathrm{x}d\mathrm{x} = [\mathrm{xlnx} -\mathrm{x}] \mathrm{from} 1 \mathrm{to} 1.99 = (1.99\times \ln 1.99 - 1.99) - ( \ln 1 -1) =0.3793 \\ \mathrm{For} {\mathrm{I}}_{2 }= {\sin }^4\mathrm{x} =({\sin }^2\mathrm{x}{)}^2 =( \frac{1-\cos 2\mathrm{x}}{2}{)}^2, \mathrm{so} \mathrm{expand} \mathrm{it}, \mathrm{and} \mathrm{integrate} \mathrm{it}. \\ \mathrm{You} \mathrm{will} \mathrm{get} \frac{3}{8}\mathrm{x} -\frac{1}{4}\sin 2\mathrm{x}+\frac{1}{32}\sin 4\mathrm{x} \\ {\int }_{1.99}^{2.8}{\sin }^4\mathrm{x}d\mathrm{x} = [\frac{3}{8}\mathrm{x} -\frac{1}{4}\sin 2\mathrm{x}+\frac{1}{32}\sin 4\mathrm{x}] \mathrm{from} 2.8 \mathrm{to} 1.99 = (1.03)-(0.71)=0.314 \\ \mathrm{So} \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{shaded} \mathrm{region} \mathrm{is} 0.379 + 0.314 =0.693 \left(\mathrm{Ans}\right) \end{gathered}$$