Calculate $\int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\sec }^2}x}}{{\sqrt[3]{{\tan x}}}}{\text{d}}x}$ .

Find $\int {{{\tan }^3}x{\text{d}}x}$ .

EITHER

let $u = \tan x;{\text{ d}}u = {\sec ^2}x{\text{d}}x$     (M1)

consideration of change of limits     (M1)

$\int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\sec }^2}x}}{{\sqrt[3]{{\tan x}}}}{\text{d}}x} = \int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{1}{{{u^{\frac{1}{3}}}}}{\text{d}}u}$     (A1)

Note: Do not penalize lack of limits.

$= \left[ {\frac{{3{u^{\frac{2}{3}}}}}{2}} \right]_1^{\sqrt 3 }$     A1

$= \frac{{3 \times {{\sqrt 3 }^{\frac{2}{3}}}}}{2} - \frac{3}{2} = \left( {\frac{{3\sqrt[3]{3} - 3}}{2}} \right)$     A1A1     N0

OR

$\int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\sec }^2}x}}{{\sqrt[3]{{\tan x}}}}{\text{d}}x} = \left[ {\frac{{3{{(\tan x)}^{\frac{2}{3}}}}}{2}} \right]_{\frac{\pi }{4}}^{\frac{\pi }{3}}$     M2A2

$= \frac{{3 \times {{\sqrt 3 }^{\frac{2}{3}}}}}{2} - \frac{3}{2} = \left( {\frac{{3\sqrt[3]{3} - 3}}{2}} \right)$     A1A1     N0

[6 marks]

$\int {{{\tan }^3}x{\text{d}}x}  = \int {\tan x({{\sec }^2}x - 1){\text{d}}x}$     M1

$= \int {(\tan x \times {{\sec }^2}x - \tan x){\text{d}}x}$

$= \frac{1}{2}{\tan ^2}x - \ln \left| {\sec x} \right| + C$     A1A1

Note: Do not penalize the absence of absolute value or C.

[3 marks]

Quite a variety of methods were successfully employed to solve part (a).

Many candidates did not attempt part (b).