By using the substitution \(t = \tan x\), find \(\int {\frac{{{\text{d}}x}}{{1 + {{\sin }^2}x}}} \).

Express your answer in the form \(m\arctan (n\tan x) + c\), where \(m\), \(n\) are constants to be determined.

EITHER

\(x = \arctan t\)     (M1)

\(\frac{{{\text{d}}x}}{{{\text{d}}t}} = \frac{1}{{1 + {t^2}}}\)     A1

OR

\(t = \tan x\)

\(\frac{{{\text{d}}t}}{{{\text{d}}x}} = {\sec ^2}x\)     (M1)

\( = 1 + {\tan ^2}x\)     A1

\( = 1 + {t^2}\)

THEN

\(\sin x = \frac{t}{{\sqrt {1 + {t^2}} }}\)     (A1)

Note:     This A1 is independent of the first two marks

\(\int {\frac{{{\text{d}}x}}{{1 + {{\sin }^2}x}} = \int {\frac{{\frac{{{\text{d}}t}}{{1 + {t^2}}}}}{{1 + {{\left( {\frac{t}{{\sqrt {1 + {t^2}} }}} \right)}^2}}}} } \)     M1A1

Note:     Award M1 for attempting to obtain integral in terms of \(t\) and \({\text{d}}t\)

\( = \int {\frac{{{\text{d}}t}}{{(1 + {t^2}) + {t^2}}} = \int {\frac{{{\text{d}}t}}{{1 + 2{t^2}}}} } \)     A1

\( = \frac{1}{2}\int {\frac{{{\text{d}}t}}{{\frac{1}{2} + {t^2}}} = \frac{1}{2} \times \frac{1}{{\frac{1}{{\sqrt 2 }}}}\arctan \left( {\frac{t}{{\frac{1}{{\sqrt 2 }}}}} \right)} \)     A1

\( = \frac{{\sqrt 2 }}{2}\arctan \left( {\sqrt 2 \tan x} \right)( + c)\)     A1

[8 marks]