Find the value of the integral $\int_0^{\sqrt 2 } {\sqrt {4 - {x^2}} {\text{d}}x}$ .

Find the value of the integral $\int_0^{0.5} {\arcsin x {\text{d}}x}$ .

Using the substitution $t = \tan \theta$ , find the value of the integral

$$\int_0^{\frac{\pi }{4}} {\frac{{{\text{d}}\theta }}{{3{{\cos }^2}\theta + {{\sin }^2}\theta }}} {\text{ }}.$$

let $x = 2\sin \theta$     M1

${\text{d}}x = 2\cos \theta {\text{d}}\theta$     A1

$I = \int_0^{\frac{\pi }{4}} {2\cos \theta \times 2\cos \theta {\text{d}}\theta \,\,\,\,\,\left( { = 4\int_0^{\frac{\pi }{4}} {{{\cos }^2}\theta {\text{d}}\theta } } \right)}$     A1A1

Note: Award A1 for limits and A1 for expression.

$= 2\int_0^{\frac{\pi }{4}} {(1 + \cos 2\theta ){\text{d}}\theta }$     A1

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

$= 1 + \frac{\pi }{2}$     A1

[7 marks]

$I = [x\arcsin x]_0^{0.5} - \int_0^{0.5} {x \times \frac{1}{{\sqrt {1 - {x^2}} }}{\text{d}}x}$     M1A1A1

$= [x\arcsin x]_0^{0.5} + \left[ {\sqrt {1 - {x^2}} } \right]_0^{0.5}$     A1

$= \frac{\pi }{{12}} + \frac{{\sqrt 3 }}{2} - 1$     A1

[5 marks]

${\text{d}}t = {\sec ^2}\theta {\text{d}}\theta {\text{ , }}\left[ {0,\frac{\pi }{4}} \right] \to [0,{\text{ 1]}}$     A1(A1)

$I = \int_0^1 {\frac{{\frac{{{\text{d}}t}}{{(1 + {t^2})}}}}{{\frac{3}{{(1 + {t^2})}} + \frac{{{t^2}}}{{(1 + {t^2})}}}}}$     M1(A1)

$= \int_0^1 {\frac{{{\text{d}}t}}{{3 + {t^2}}}}$     A1

$= \frac{1}{{\sqrt 3 }}\left[ {\arctan \left( {\frac{x}{{\sqrt 3 }}} \right)} \right]_0^1$     A1

$= \frac{\pi }{{6\sqrt 3 }}$     A1

[7 marks]