In(x)=∫0x1(t2+5)ndt,n=1,2,3,⋯\begin{array}{l} I_n\left(x\right)=\int_0^x\frac{1}{\left(t^2+5\right)^n}dt, n=1, 2, 3,\cdots\end{array}In(x)=∫0x(t2+5)n1dt,n=1,2,3,⋯
Then
If dydx+2ytanx=sinx,0\frac{d y}{d x}+2 y \tan x=\sin x, 0dxdy+2ytanx=sinx,0