Using the substitution \(x = \tan \theta \) show that \(\int\limits_0^1 {\frac{1}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{d}}x = } \int\limits_0^{\frac{\pi }{4}} {{{\cos }^2}\theta {\text{d}}\theta } \).

Hence find the value of \(\int\limits_0^1 {\frac{1}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{d}}x} \).

let \(x = \tan \theta \)

\( \Rightarrow \frac{{{\text{d}}x}}{{{\text{d}}\theta }} = {\sec ^2}\theta \)     (A1)

\(\int {\frac{1}{{{{({x^2} + 1)}^2}}}{\text{d}}x = \int {\frac{{{{\sec }^2}\theta }}{{{{({{\tan }^2}\theta + 1)}^2}}}{\text{d}}\theta } } \)     M1

Note:     The method mark is for an attempt to substitute for both \(x\) and \({\text{d}}x\).

\( = \int {\frac{1}{{{{\sec }^2}\theta }}{\text{d}}\theta } \) (or equivalent)     A1

when \(x = 0,{\text{ }}\theta = 0\) and when \(x = 1,{\text{ }}\theta = \frac{\pi }{4}\)     M1

\(\int\limits_0^{\frac{\pi }{4}} {{{\cos }^2}\theta {\text{d}}\theta } \)    AG

[4 marks]

\(\left( {\int\limits_0^1 {\frac{1}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{d}}x}  = \int\limits_0^{\frac{\pi }{4}} {{{\cos }^2}\theta {\text{d}}\theta } } \right) = \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\left( {1 + \cos 2\theta } \right){\text{d}}\theta } \)   M1

\( = \frac{1}{2}\left[ {\theta + \frac{{\sin 2\theta }}{2}} \right]_0^{\frac{\pi }{4}}\)     A1

\( = \frac{\pi }{8} + \frac{1}{4}\)     A1

[3 marks]