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 \(a\) .
▶️Answer/Explanation
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.
▶️Answer/Explanation
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 \(\int {\frac{{{x^3}}}{{\sqrt {1 – {x^2}} }}{\text{d}}x} \) .
▶️Answer/Explanation
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 \(\int {\frac{{{\text{d}}x}}{{x\sqrt {{x^4} – 4} }} = \frac{1}{4}\arccos \left( {\frac{2}{{{x^2}}}} \right) + c} \).
▶️Answer/Explanation
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]