Question

If integral 02 a 1 upon 4 + x squared dx is equal to pi by 8 then the value of a is

Answer 1

The value of 'a' is 2.

• Given,

                $\int\limits^a_0 {\frac{1}{4+x^{2} } } \, dx =\frac{\pi }{8}$

                $\int\limits^a_0 {\frac{1}{2^{2} +x^{2} } } \, dx =\frac{\pi }{8}$

                $[\frac{1}{2}tan^{-1}(\frac{x}{2})]^{a}_{0}=\frac{\pi }{8}$              [∵ $\int\ {\frac{1}{x^{2}+a^{2} } } \, dx =\frac{1}{a}tan^{-1}(\frac{x}{a})+c$]

                $tan^{-1}(\frac{a}{2})-tan^{-1}(\frac{0}{2})=\frac{\pi }{4}$

                $tan^{-1}(\frac{a}{2})=\frac{\pi }{4}$

                 $(\frac{a}{2})=tan\frac{\pi }{4}$

                  $\frac{a}{2}=1$

                ∴ a = 2