Home / IBDP Maths AA: Topic: SL 5.5: Introduction to integration as anti-differentiation: IB style Questions HL Paper 2

IBDP Maths AA: Topic: SL 5.5: Introduction to integration as anti-differentiation: IB style Questions HL Paper 2

Question

By using the substitution \(x = 2\tan u\), show that \(\int {\frac{{{\text{d}}x}}{{{x^2}\sqrt {{x^2} + 4} }} = \frac{{ – \sqrt {{x^2} + 4} }}{{4x}} + C} \).

▶️Answer/Explanation

Markscheme

EITHER

\(\frac{{{\text{d}}x}}{{{\text{d}}u}} = 2\,{\text{se}}{{\text{c}}^2}u\)     A1

\(\int {\frac{{2\,{\text{se}}{{\text{c}}^2}u{\text{d}}u}}{{4{{\tan }^2}u\sqrt {4 + 4{{\tan }^2}u} }}} \)     (M1)

\(\int {\frac{{2\,{\text{se}}{{\text{c}}^2}u{\text{d}}u}}{{4{{\tan }^2}u \times 2\,\sec u}}} \)   \(( = \int {\frac{{{\text{d}}u}}{{4{{\sin }^2}u\sqrt {{{\tan }^2}u + 1} }}{\text{ or }} = \int {\frac{{2\,{\text{se}}{{\text{c}}^2}u{\text{d}}u}}{{4{{\tan }^2}u\sqrt {4{{\sec }^2}u} }})} } \)     A1

OR

\(u = \arctan \frac{x}{2}\)

\(\frac{{{\text{d}}u}}{{{\text{d}}x}} = \frac{2}{{{x^2} + 4}}\)     A1

\(\int {\frac{{\sqrt {4{{\tan }^2}u + 4{\text{d}}u} }}{{2 \times 4{{\tan }^2}u}}} \)     (M1)

\(\int {\frac{{2\,\sec u{\text{d}}u}}{{2 \times 4{{\tan }^2}u}}} \)     A1

THEN

\( = \frac{1}{4}\int {\frac{{\sec u{\text{d}}u}}{{{{\tan }^2}u}}} \)

\( = \frac{1}{4}\int {{\text{cosec}}\,u\cot u{\text{d}}u{\text{ }}\left( { = \frac{1}{4}\int {\frac{{\cos u}}{{{{\sin }^2}u}}{\text{d}}u} } \right)} \)     A1

\( =  – \frac{1}{4}{\text{cosec}}\,u( + C){\text{ }}\left( { =  – \frac{1}{{4\sin u}}( + C)} \right)\)     A1

use of either \(u = \frac{x}{2}\) or an appropriate trigonometric identity     M1

either \(\sin u = \frac{x}{{\sqrt {{x^2} + 4} }}\) or \({\text{cosec}}\,u = \frac{{\sqrt {{x^2} + 4} }}{x}\) (or equivalent)     A1

\( = \frac{{ – \sqrt {{x^2} + 4} }}{{4x}}( + C)\)     AG

[7 marks]

Examiners report

Most candidates found this a challenging question. A large majority of candidates were able to change variable from x to u but were not able to make any further progress.

Question

a.Given that \(2{x^3} – 3x + 1\) can be expressed in the form \(Ax\left( {{x^2} + 1} \right) + Bx + C\), find the values of the constants \(A\), \(B\) and \(C\).[2]

b.Hence find \(\int {\frac{{2{x^3} – 3x + 1}}{{{x^2} + 1}}} {\text{d}}x\).[5]

▶️Answer/Explanation

Markscheme

\(2{x^3} – 3x + 1 = Ax\left( {{x^2} + 1} \right) + Bx + C\)

\(A = 2,\,\,C = 1,\)     A1

\(A + B =  – 3 \Rightarrow B =  – 5\)     A1

[2 marks]

a.

\(\int {\frac{{2{x^3} – 3x + 1}}{{{x^2} + 1}}} {\text{d}}x = \int {\left( {2x – \frac{{5x}}{{{x^2} + 1}} + \frac{1}{{{x^2} + 1}}} \right)} {\text{d}}x\)      M1M1

Note: Award M1 for dividing by \(\left( {{x^2} + 1} \right)\) to get \(2x\), M1 for separating the \(5x\) and 1.

\( = {x^2} – \frac{5}{2}{\text{ln}}\left( {{x^2} + 1} \right) + {\text{arctan}}\,x\left( { + c} \right)\)     (M1)A1A1

Note: Award (M1)A1 for integrating \({\frac{{5x}}{{{x^2} + 1}}}\), A1 for the other two terms.

[5 marks]

 

 

 

 
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