Question

The value of the integral \(\displaystyle\int_0^{2\pi}\left(\dfrac{3}{9+\sin^2 \theta }\right)d\theta \) is
1. \(\dfrac{2\pi}{\sqrt{10}}\)
2. \(2\sqrt{10} \pi\)
3. \(\sqrt{10} \pi\)
4. π 

Answer 1

Correct Answer - Option 1 : \(\dfrac{2\pi}{\sqrt{10}}\)

\(\mathop \smallint \limits_0^{2a} f\left( x \right)dx = \left\{ {\begin{array}{*{20}{c}} {2\mathop \smallint \limits_0^a f\left( x \right)dx,\;\;f\left( {2a - x} \right) = f\left( x \right)}\\ {0,\;\;f\left( {2a - x} \right) = - f\left( x \right)} \end{array}} \right.\)

From the property of integration as mentioned above, the given integral can be reduced to

\( = 2\mathop \smallint \limits_0^\pi \left( {\frac{3}{{9 + {{\sin }^2}\theta }}} \right)d\theta \)

\( = 4\mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {\frac{3}{{9 + {{\sin }^2}\theta }}} \right)d\theta \)

\( = 4\mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {\frac{3}{{9 + \frac{{{{\cos }^2}\theta }}{{{{\cos }^2}\theta }}{{\sin }^2}\theta }}} \right)d\theta \)

\( = 12\mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {\frac{{{{\sec }^2}\theta }}{{9{{\sec }^2}\theta + {{\tan }^2}\theta }}} \right)d\theta \)

\( = 12\mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {\frac{{{{\sec }^2}\theta }}{{9\left( {1 + {{\tan }^2}\theta } \right) + {{\tan }^2}\theta }}} \right)d\theta \)

\( = 12\mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {\frac{{{{\sec }^2}\theta }}{{9 + 10{{\tan }^2}\theta }}} \right)d\theta \)

Put tan θ = t

⇒ sec²θ dθ = dt

\( = 12\mathop \smallint \limits_0^\infty \left( {\frac{1}{{9 + 10{t^2}}}} \right)dt\)

\( = \frac{{12}}{{10}}\mathop \smallint \limits_0^\infty \left( {\frac{1}{{\frac{9}{{10}} + {t^2}}}} \right)dt\)

\( = \frac{{12}}{{10}} \times \frac{1}{{\frac{3}{{\sqrt {10} }}}}\left[ {{{\tan }^{ - 1}}\left( {\frac{t}{{\frac{3}{{\sqrt {10} }}}}} \right)} \right]_0^\infty \)

\( = \frac{4}{{\sqrt {10} }}\left[ {\frac{\pi }{2} - 0} \right] = \frac{{2\pi }}{{\sqrt {10} }}\)