Practicing Success
$\int \frac{dx}{\sqrt{5-x}}$ = (where C is arbitrary constant.) |
$\sqrt{5-x}+C$ $-\sqrt{5-x}+C$ $2 \sqrt{5-x}+C$ $-2 \sqrt{5-x}+C$ |
$-2 \sqrt{5-x}+C$ |
$I=\int \frac{1}{\sqrt{5-x}} d x$ .....(1) let $y=5-x$ So $d y=-d x \Rightarrow d x=-d y$ substituting y in eq. 1 $I =\int \frac{-d y}{\sqrt{y}}$ $=-\int y^{-1 / 2} d y$ $=\frac{-y^{-1 / 2+1}}{-1 / 2+1}+C=\frac{-y^{1 / 2}}{1 / 2}+C$ $=-2 \sqrt{y}+C$ substituting y in terms of x C → arbitrary constant $-2 \sqrt{5-x}+c$ |