Question
Calculate \(\int_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\sec }^2}x}}{{\sqrt[3]{{\tan x}}}}{\text{d}}x} \) .
Find \(\int {{{\tan }^3}x{\text{d}}x} \) .
▶️Answer/Explanation
Markscheme
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]
Examiners report
Quite a variety of methods were successfully employed to solve part (a).
Many candidates did not attempt part (b).