Question

Solve the integral

$\int \sqrt{3-x}\sqrt{x-1}dx$

Short answer

$\frac{{\sin }^{-}1(x-2)}{2}+\frac{1}{4}sin\left(2si{n}^{-}1\right(x-2\left)\right)+C$

Step-by-step solution

Step 1. Given

The given function is $\int \sqrt{3-x}\sqrt{x-1}dx$

Step 2. Simplified the integral

$$\begin{gathered} \mathrm{The}\mathrm{integral}\mathrm{is}\mathrm{simplified}\mathrm{below}. \\ \int \sqrt{3-x)}\sqrt{(x-1)}dx \\ =\int \sqrt{-x^2+4x-3}dx \\ =\int \sqrt{-(x-2{)}^2+1}dx \\ =\int \sqrt{-u^2+1}dx[u=x-2,du=dx] \end{gathered}$$

Step 3.  Using the identity

$$\begin{gathered} \mathrm{To}\mathrm{solve}\mathrm{the}\mathrm{integral},\mathrm{le}tx=sin\left(v\right). \\ \mathrm{Now},\mathrm{derivation}\mathrm{of}\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{is}\mathrm{solved}\mathrm{below}. \\ u=sin\left(v\right)du \\ =cos\left(v\right)dv \\ \mathrm{Now},\mathrm{use}\mathrm{the}\mathrm{identity} \\ -si{n}^2\left(u\right)+1=co{s}^2\left(u\right) \\ So,\sqrt{(-u^2+1)}=\sqrt{(-{\sin }^2(v)+1}) \\ =\sqrt{{\cos }^2\left(v\right)}=cos(y) \end{gathered}$$

Step 4. Calculation

$$\begin{gathered} \mathrm{The}\mathrm{integral}\mathrm{is}\mathrm{solved}\mathrm{below}. \\ \surd =u²+1dx \\ =\int \cos ²\left(v\right)dv \\ =\int \left(\frac{1}{2}\cos \left(2v\right)+\frac{1}{2}\right)dv \\ =\int \frac{1}{2}dv-\int \left(\frac{1}{2}\cos \left(2v\right)dv\right) \\ =\frac{v}{2}+\frac{1}{2}\int \frac{1}{2}cos\left(w\right)dw[w=2v,dw=2dv] \\ =\frac{v}{2}+\frac{1}{4}sin\left(w\right)+C \\ =\frac{v}{2}+\frac{1}{4}sin\left(2v\right)+C \\ =\frac{{\sin }^{-}1\left(u\right)}{2}+\frac{1}{4}sin\left(2si{n}^{-}1\right(u\left)\right)+C \\ =\frac{{\sin }^{-}1(x-2)}{2}+\frac{1}{4}sin\left(2si{n}^{-}1\right(x-2\left)\right)+C \end{gathered}$$