5.11 Integrals with square roots in denominator

We shall consider only one type

The coefficient of {x}^{2} must be negative, if it is positive we need a different approach which involves hyperbolic functions (not discussed here). The method is as follows

1. Complete the square, a + bx − {x}^{2} = {d}^{2} −{ (x + c)}^{2}, with c = −b∕2 and {d}^{2} = a + {b}^{2}∕4.
2. Substitute z = x − c, which gives us the derivative of the \mathop{arcsin}\nolimits .

Example 5.25: 

Calculate I =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits {dx\over \sqrt{3 + 4x − {x}^{2}}}.

Solution: 

Complete the square: 3 + 4x − {x}^{2} = {d}^{2} −{ (x + c)}^{2}. Equate the coefficients of each power. {x}^{2}: − 1 = −1, contains no unknowns. x: 4 = −2c (therefore c = −2). The constant term gives 3 = {d}^{2} − {c}^{2} = {d}^{2} − 4, and thus {d}^{2} = 7, d = \sqrt{7}, and

I =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits {dx\over \sqrt{7 − {(x − 2)}^{2}}}.

We substitute z = x − 2, dz = dz∕dx\kern 1.66702pt dx = dx, which leads to

\class{boxed}{I =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits {dz\over \sqrt{{\sqrt{7} }^{2 } − {z}^{2}}} ={\mathop{ sin}\nolimits }^{−1} {z\over \sqrt{7}} + k ={\mathop{ sin}\nolimits }^{−1}\left ({x − 2\over \sqrt{7}} \right ) + k. }

(The integral is a standard integral and can be found in the tables, but is easily checked by using the chain rule and

{d\over dy}{\mathop{sin}\nolimits }^{−1}y = {1\over \sqrt{1 − {y}^{2}}}.\quad

)