Question

Evaluate the Integral\(\int\limits_0^a {\frac{{dx}}{{{{\left( {{a^2} + {b^2}} \right)}^{3/2}}}}} \) , a>0

Short answer

The definite integration of \(\int\limits_0^a {\frac{{dx}}{{{{\left( {{a^2} + {b^2}} \right)}^{3/2}}}}} \) is \(\frac{1}{{\sqrt 2 {a^2}}}\).

Step-by-step solution

Step (1):- Convert “dx” in terms of .

Let \(x = a\tan \theta \) (differentiate on both sides)

\( \Rightarrow dx = a{\sec ^2}\theta d\theta \) ( since, \(d(\tan \theta ) = {\sec ^2}\theta d\theta \))

Step (2):- Substitute x and dx in given equation

\(\int\limits_0^{\pi /4} {\frac{{a{{\sec }^2}\theta d\theta }}{{{{\left( {{a^2} + {{\left( {a\tan \theta } \right)}^2}} \right)}^{3/2}}}}} \)( since, \( - \frac{\pi }{2} < \theta < \frac{\pi }{2}\))

\(\int\limits_0^{\pi /4} {\frac{{a{{\sec }^2}\theta }}{{{{\left( {{a^2} + {{\left( {a\tan \theta } \right)}^2}} \right)}^{3/2}}}}} .d\theta \)

\(\int\limits_0^{\pi /4} {\frac{{a{{\sec }^2}\theta }}{{{{\left( {{a^3}{{\left( {1 + {{\tan }^2}\theta } \right)}^{3/2}}} \right)}^{}}}}.d\theta } \)

\(\int\limits_0^{\pi /4} {\left( {\frac{1}{{{a^4}}}} \right).\left( {\frac{{{{\sec }^2}\theta }}{{{{\left( {1 + {{\tan }^2}\theta } \right)}^{3/2}}}}} \right).d\theta } \)

\(\left( {\frac{1}{{{a^2}}}} \right)\int\limits_0^{\pi /4} {\left( {\frac{{{{\sec }^2}\theta }}{{{{\left( {{{\sec }^2}\theta } \right)}^{3/2}}}}} \right).d\theta = } \left( {\frac{1}{{{a^2}}}} \right)\int\limits_0^{\pi /4} {\left( {\frac{{{{\sec }^2}\theta }}{{\left( {{{\sec }^3}\theta } \right)}}} \right).d\theta } \)

\(\left( {\frac{1}{{{a^2}}}} \right)\int\limits_0^{\pi /4} {\frac{1}{{\sec \theta }}.d\theta } \)

\(\int\limits_0^a {\frac{{dx}}{{{{\left( {{a^2} + {x^2}} \right)}^{3/2}}}} = \int\limits_0^{\pi /4} {\frac{{a{{\sec }^2}\theta d\theta }}{{{{\left( {{a^2} + {{\left( {a\tan \theta } \right)}^2}} \right)}^{3/2}}}}} } \)

\(\begin{aligned}{l} &= \int\limits_0^{\pi /4} {\frac{1}{{\sec \theta }}d\theta .\left( {\frac{1}{{{a^2}}}} \right)} \\ &= \int\limits_0^{\pi /4} {\cos \theta d\theta .\left( {\frac{1}{{{a^2}}}} \right)} \end{aligned}\)

Since, \(\frac{1}{{\sec \theta }} = \cos \theta \)

Step (3):- Integrate \(\cos \theta \) between 0 to \(\frac{\pi }{4}.\)

\( = \int\limits_0^a {\frac{{dx}}{{{{\left( {{a^2} + {x^2}} \right)}^{3/2}}}} = \int\limits_0^{\pi /4} {\cos \theta d\theta .\left( {\frac{1}{{{a^2}}}} \right)} } \)

=\(\left( {\frac{1}{{{a^2}}}} \right)\left( {\sin \theta } \right)_0^{\pi /4}\)

Since, \(\int {\cos \theta = \sin \theta } \)

\(\int\limits_0^a {\frac{{dx}}{{{{\left( {{a^2} + {x^2}} \right)}^{3/2}}}} = \left( {\sin \frac{\pi }{4} - \sin 0} \right)\left( {\frac{1}{{{a^2}}}} \right)} \)

=\(\left( {\frac{1}{{\sqrt 2 }} - 0} \right)\left( {\frac{1}{{{a^2}}}} \right)\)

\(\int\limits_0^a {\frac{{dx}}{{{{\left( {{a^2} + {x^2}} \right)}^{3/2}}}} = \frac{1}{{\sqrt 2 {a^2}}}} \)

Hence, the definite integration of \(\int\limits_0^a {\frac{{dx}}{{{{\left( {{a^2} + {x^2}} \right)}^{3/2}}}} = \frac{1}{{\sqrt 2 {a^2}}}} \).