# IB DP Math AA Topic: AHL 5.16 Integration by substitution: IB Style Questions HL Paper 2

## Question

(a)     Integrate $$\int {\frac{{\sin \theta }}{{1 – \cos \theta }}} {\text{d}}\theta$$ .

(b)     Given that $$\int_{\frac{\pi }{2}}^a {\frac{{\sin \theta }}{{1 – \cos \theta }}} {\text{d}}\theta = \frac{1}{2}$$ and $$\frac{\pi }{2} < a < \pi$$, find the value of $$a$$ .

## Markscheme

(a)     $$\int {\frac{{\sin \theta }}{{1 – \cos \theta }}} {\text{d}}\theta = \int {\frac{{\left( {1 – \cos \theta } \right)’}}{{1 – \cos \theta }}} {\text{d}}\theta = \ln \left( {1 – \cos \theta } \right) + C$$     (M1)A1A1

Note: Award A1 for $$\ln \left( {1 – \cos \theta } \right)$$ and A1 for C.

(b)     $$\int_{\frac{\pi }{2}}^a {\frac{{\sin \theta }}{{1 – \cos \theta }}} {\text{d}}\theta = \frac{1}{2} \Rightarrow \left[ {\ln \left( {1 – \cos \theta } \right)} \right]_{\frac{\pi }{2}}^a = \frac{1}{2}$$     M1

$$1 – \cos a = {{\text{e}}^{\frac{1}{2}}} \Rightarrow a = \arccos \left( {1 – \sqrt {\text{e}} } \right)$$) or $$2.28$$     A1     N2

[5 marks]

## Examiners report

Generally well answered, although many students did not include the constant of integration.

## Question

Using the substitution $$x = 2\sin \theta$$ , show that$\int {\sqrt {4 – {x^2}} } {\text{d}}x = Ax\sqrt {4 – {x^2}} + B\arcsin \frac{x}{2} + {\text{constant ,}}$where $$A$$ and $$B$$ are constants whose values you are required to find.

## Markscheme

$$\int {\sqrt {4 – {x^2}} } {\text{d}}x$$

$$x = 2\sin \theta$$

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

$$= \int {\sqrt {4 – 4{{\sin }^2}\theta } } \times 2\cos \theta {\text{d}}\theta$$     M1A1

$$= \int {2\cos \theta \times } 2\cos \theta {\text{d}}\theta$$

$$= 4\int {{{\cos }^2}\theta {\text{d}}\theta }$$

now $$\int {{{\cos }^2}\theta {\text{d}}\theta }$$

$$= \int {\left( {\frac{1}{2}\cos 2\theta + \frac{1}{2}} \right)} {\text{d}}\theta$$     M1A1

$$= \left( {\frac{{\sin 2\theta }}{4} + \frac{1}{2}\theta } \right)$$    A1

so original integral

$$= 2\sin 2\theta + 2\theta$$

$$= 2\sin \theta \cos \theta + 2\theta$$

$$= \left( {2 \times \frac{x}{2} \times \frac{{\sqrt {4 – {x^2}} }}{2}} \right) + 2\arcsin \left( {\frac{x}{2}} \right)$$

$$= \frac{{x\sqrt {4 – {x^2}} }}{2} + 2\arcsin \left( {\frac{x}{2}} \right) + C$$     A1A1

Note: Do not penalise omission of $$C$$.

$$\left( {A = \frac{1}{2},{\text{ }}B = 2} \right)$$

[8 marks]

## Examiners report

For many candidates this was an all or nothing question. Examiners were surprised at the number of candidates who were unable to change the variable in the integral using the given substitution. Another stumbling block, for some candidates, was a lack of care with the application of the trigonometric version of Pythagoras’ Theorem to reduce the integrand to a multiple of $${\cos ^2}\theta$$ . However, candidates who obtained the latter were generally successful in completing the question.

## Question

By using the substitution $$x = \sin t$$ , find $$\int {\frac{{{x^3}}}{{\sqrt {1 – {x^2}} }}{\text{d}}x}$$ .

## Markscheme

$$x = \sin t,{\text{ d}}x = \cos t\,{\text{d}}t$$

$$\int {\frac{{{x^3}}}{{\sqrt {1 – {x^2}} }}{\text{d}}x} = \int {\frac{{{{\sin }^3}t}}{{\sqrt {1 – {{\sin }^2}t} }}\cos t\,{\text{d}}t}$$     M1

$$= \int {{{\sin }^3}t\,{\text{d}}t}$$     (A1)

$$= \int {{{\sin }^2}t\sin t\,{\text{d}}t}$$

$$= \int {(1 – {{\cos }^2}t)\sin t\,{\text{d}}t}$$     M1A1

$$= \int {\sin t\,{\text{d}}t – \int {{{\cos }^2}t\sin t\,{\text{d}}t} }$$

$$= – \cos t + \frac{{{{\cos }^3}t}}{3} + C$$     A1A1

$$= – \sqrt {1 – {x^2}} + \frac{1}{3}{\left( {\sqrt {1 – {x^2}} } \right)^3} + C$$     A1

$$\left( { = – \sqrt {1 – {x^2}} \left( {1 – \frac{1}{3}(1 – {x^2})} \right) + C} \right)$$

$$\left( { = – \frac{1}{3}\sqrt {1 – {x^2}} (2 + {x^2}) + C} \right)$$

[7 marks]

## Examiners report

Just a few candidates got full marks in this question. Substitution was usually incorrectly done and lead to wrong results. A cosine term in the denominator was a popular error. Candidates often chose unhelpful trigonometric identities and attempted integration by parts. Results such as $$\int {{{\sin }^3}t\,{\text{d}}t = \frac{{{{\sin }^4}t}}{4} + C}$$ were often seen along with other misconceptions concerning the manipulation/simplification of integrals were also noticed. Some candidates unsatisfactorily attempted to use $$\arcsin x$$ . However, there were some good solutions involving an expression for the cube of $$\sin t$$ in terms of $$\sin t$$ and $$\sin 3t$$ . Very few candidates re-expressed their final result in terms of x.

## Question

By using the substitution $${x^2} = 2\sec \theta$$, show that $$\int {\frac{{{\text{d}}x}}{{x\sqrt {{x^4} – 4} }} = \frac{1}{4}\arccos \left( {\frac{2}{{{x^2}}}} \right) + c}$$.

## Markscheme

EITHER

$${x^2} = 2\sec \theta$$

$$2x\frac{{{\text{d}}x}}{{{\text{d}}\theta }} = 2\sec \theta \tan \theta$$     M1A1

$$\int {\frac{{{\text{d}}x}}{{x\sqrt {{x^4} – 4} }}}$$

$$= \int {\frac{{\sec \theta \tan \theta {\text{d}}\theta }}{{2\sec \theta \sqrt {4{{\sec }^2}\theta – 4} }}}$$     M1A1

OR

$$x = \sqrt 2 {(\sec \theta )^{\frac{1}{2}}}{\text{ }}\left( { = \sqrt 2 {{(\cos \theta )}^{ – \frac{1}{2}}}} \right)$$

$$\frac{{{\text{d}}x}}{{{\text{d}}\theta }} = \frac{{\sqrt 2 }}{2}{(\sec \theta )^{\frac{1}{2}}}\tan \theta {\text{ }}\left( { = \frac{{\sqrt 2 }}{2}{{(\cos \theta )}^{ – \frac{3}{2}}}\sin \theta } \right)$$     M1A1

$$\int {\frac{{{\text{d}}x}}{{x\sqrt {{x^4} – 4} }}}$$

$$= \int {\frac{{\sqrt 2 {{(\sec \theta )}^{\frac{1}{2}}}\tan \theta {\text{d}}\theta }}{{2\sqrt 2 {{(\sec \theta )}^{\frac{1}{2}}}\sqrt {4{{\sec }^2}\theta – 4} }}{\text{ }}\left( { = \int {\frac{{\sqrt 2 {{(\cos \theta )}^{ – \frac{3}{2}}}\sin \theta {\text{d}}\theta }}{{2\sqrt 2 {{(\cos \theta )}^{ – \frac{1}{2}}}\sqrt {4{{\sec }^2}\theta – 4} }}} } \right)}$$     M1A1

THEN

$$= \frac{1}{2}\int {\frac{{\tan \theta {\text{d}}\theta }}{{2\tan \theta }}}$$     (M1)

$$= \frac{1}{4}\int {{\text{d}}\theta }$$

$$= \frac{\theta }{4} + c$$     A1

$${x^2} = 2\sec \theta \Rightarrow \cos \theta = \frac{2}{{{x^2}}}$$     M1

Note:     This M1 may be seen anywhere, including a sketch of an appropriate triangle.

so $$\frac{\theta }{4} + c = \frac{1}{4}\arccos \left( {\frac{2}{{{x^2}}}} \right) + c$$     AG

[7 marks]

[N/A]
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