# IBDP Maths analysis and approaches Topic: SL 5.5 Introduction to integration as anti-differentiation of functions HL Paper 1

## Question

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

[6]
a.

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

[3]
b.

## 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]

a.

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

b.

## Examiners report

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

a.

Many candidates did not attempt part (b).

b.
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