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]