IBDP Maths AA: Topic SL 5.10: Indefinite integral: IB style Questions SL Paper 1

Question

The graph of a function f passes through the point (ln4, 20).

Given that f ′(x) = 6e2x , find f (x).

Answer/Explanation

Ans:

evidence of integration

eg \(\int f'(x)dx, \int 6e^{2x}\)

correct integration (accept missing + c)

eg \(\frac{1}{2}\times 6e^{2x}, 3e^{2x}\) +c

substituting initial condition into their integrated expression (must have +c )

eg \(3e^{2\times ln4}\)+c = 20

correct application of log (\(a^{b})\)= b log a rule (seen anywhere)

eg 2ln4 = ln16, eln16, ln42

correct application of  elna= a rule (seen anywhere)

eg \(e^{ln16}= 16, (e^{ln4})^{2}=4^{2}\)

correct working

eg \(3\times 16+c= 20, 3\times (4^{2})+c= 20, c=20 c=-28\)

\(f(x)=3e^{2x}- 28\)

Question

Let \(f(x) = \int {\frac{{12}}{{2x – 5}}} {\rm{d}}x\) , \(x > \frac{5}{2}\) . The graph of \(f\) passes through (\(4\), \(0\)) .

Find \(f(x)\) .

Answer/Explanation

Markscheme

attempt to integrate which involves \(\ln \)     (M1)

eg   \(\ln (2x – 5)\) , \(12\ln 2x – 5\) , \(\ln 2x\)

correct expression (accept absence of \(C\))

eg   \(12\ln (2x – 5)\frac{1}{2} + C\) , \(6\ln (2x – 5)\)     A2

attempt to substitute (4,0) into their integrated     (M1)

eg \(0 = 6\ln (2 \times 4 – 5)\) , \(0 = 6\ln (8 – 5) + C\)

\(C = – 6\ln 3\)     (A1)

\(f(x) = 6\ln (2x – 5) – 6\ln 3\) \(\left( { = 6\ln \left( {\frac{{2x – 5}}{3}} \right)} \right)\) (accept \(6\ln (2x – 5) – \ln {3^6}\) )     A1     N5

Note: Exception to the FT rule. Allow full FT on incorrect integration which must involve \(\ln\).

[6 marks]

Question

A function f (x) has derivative f ′(x) = 3x2 + 18x. The graph of f has an x-intercept at x = −1.

Find f (x).

[6]
a.

The graph of f has a point of inflexion at x = p. Find p.

[4]
b.

Find the values of x for which the graph of f is concave-down.

[3]
c.
Answer/Explanation

Markscheme

evidence of integration       (M1)

eg  \(\int {f’\left( x \right)} \)

correct integration (accept absence of C)       (A1)(A1)

eg  \({x^3} + \frac{{18}}{2}{x^2} + C,\,\,{x^3} + 9{x^2}\)

attempt to substitute x = −1 into their = 0 (must have C)      M1

eg  \({\left( { – 1} \right)^3} + 9{\left( { – 1} \right)^2} + C = 0,\,\, – 1 + 9 + C = 0\)

Note: Award M0 if they substitute into original or differentiated function.

correct working       (A1)

eg  \(8 + C = 0,\,\,\,C =  – 8\)

\(f\left( x \right) = {x^3} + 9{x^2} – 8\)      A1 N5

[6 marks]

a.

METHOD 1 (using 2nd derivative)

recognizing that f” = 0 (seen anywhere)      M1

correct expression for f”      (A1)

eg   6x + 18, 6p + 18

correct working      (A1)

6+ 18 = 0

p = −3       A1 N3

METHOD 1 (using 1st derivative)

recognizing the vertex of f′ is needed       (M2)

eg   \( – \frac{b}{{2a}}\) (must be clear this is for f′)

correct substitution      (A1)

eg   \(\frac{{ – 18}}{{2 \times 3}}\)

p = −3       A1 N3

[4 marks]

b.

valid attempt to use f” (x) to determine concavity      (M1)

eg   f” (x) < 0, f” (−2), f” (−4),  6x + 18 ≤ 0 

correct working       (A1)

eg   6x + 18 < 0, f” (−2) = 6, f” (−4) = −6 

f concave down for x < −3 (do not accept ≤ −3)       A1 N2

[3 marks]

c.

Question

Let \(f\left( x \right) = 6{x^2} – 3x\). The graph of \(f\) is shown in the following diagram.

Find \(\int {\left( {6{x^2} – 3x} \right){\text{d}}x} \).

[2]
a.

Find the area of the region enclosed by the graph of \(f\), the x-axis and the lines x = 1 and x = 2 .

[4]
b.
Answer/Explanation

Markscheme

\(2{x^3} – \frac{{3{x^2}}}{2} + c\,\,\,\left( {{\text{accept}}\,\,\frac{{6{x^3}}}{3} – \frac{{3{x^2}}}{2} + c} \right)\)     A1A1 N2

Notes: Award A1A0 for both correct terms if +c is omitted.
Award A1A0 for one correct term eg \(2{x^3} + c\).
Award A1A0 if both terms are correct, but candidate attempts further working to solve for c.

[2 marks]

a.

substitution of limits or function (A1)

eg  \(\int_1^2 {f\left( x \right)} \,{\text{d}}x,\,\,\left[ {2{x^3} – \frac{{3{x^2}}}{2}} \right]_1^2\)

substituting limits into their integrated function and subtracting     (M1)

eg  \(\frac{{6 \times {2^3}}}{3} – \frac{{3 \times {2^2}}}{2} – \left( {\frac{{6 \times {1^3}}}{3} + \frac{{3 \times {1^2}}}{2}} \right)\)

Note: Award M0 if substituted into original function.

correct working      (A1)

eg  \(\frac{{6 \times 8}}{3} – \frac{{3 \times 4}}{2} – \frac{{6 \times 1}}{3} + \frac{{3 \times 1}}{2},\,\,\left( {16 – 6} \right) – \left( {2 – \frac{3}{2}} \right)\)

\(\frac{{19}}{2}\)     A1 N3

[4 marks]

b.

Question

Given that \(\int_0^5 {\frac{2}{{2x + 5}}} {\rm{d}}x = \ln k\) , find the value of k .

Answer/Explanation

Markscheme

correct integration, \(2 \times \frac{1}{2}\ln (2x + 5)\)     A1A1

Note: Award A1 for \(2 \times \frac{1}{2}( = 1)\) and A1 for \(\ln (2x + 5)\) .

evidence of substituting limits into integrated function and subtracting     (M1)

e.g. \(\ln (2 \times 5 + 5) – \ln (2 \times 0 + 5)\)

correct substitution     A1

e.g. \(\ln 15 – \ln 5\)

correct working     (A1)

e.g. \(\ln \frac{{15}}{5},\ln 3\)

\(k = 3\)     A1     N3 

[6 marks]

Question

Let \(f(x) = \int {\frac{{12}}{{2x – 5}}} {\rm{d}}x\) , \(x > \frac{5}{2}\) . The graph of \(f\) passes through (\(4\), \(0\)) .

Find \(f(x)\) .

Answer/Explanation

Markscheme

attempt to integrate which involves \(\ln \)     (M1)

eg   \(\ln (2x – 5)\) , \(12\ln 2x – 5\) , \(\ln 2x\)

correct expression (accept absence of \(C\))

eg   \(12\ln (2x – 5)\frac{1}{2} + C\) , \(6\ln (2x – 5)\)     A2

attempt to substitute (4,0) into their integrated     (M1)

eg \(0 = 6\ln (2 \times 4 – 5)\) , \(0 = 6\ln (8 – 5) + C\)

\(C = – 6\ln 3\)     (A1)

\(f(x) = 6\ln (2x – 5) – 6\ln 3\) \(\left( { = 6\ln \left( {\frac{{2x – 5}}{3}} \right)} \right)\) (accept \(6\ln (2x – 5) – \ln {3^6}\) )     A1     N5

Note: Exception to the FT rule. Allow full FT on incorrect integration which must involve \(\ln\).

[6 marks]

Question

Let  \(f:x \mapsto {\sin ^3}x\) .

(i) Write down the range of the function f .

(ii) Consider \(f(x) = 1\) , \(0 \le x \le 2\pi \) . Write down the number of solutions to this equation. Justify your answer.

[5]
a.

Find \(f'(x)\) , giving your answer in the form \(a{\sin ^p}x{\cos ^q}x\) where \(a{\text{, }}p{\text{, }}q \in \mathbb{Z}\) .

[2]
b.

Let \(g(x) = \sqrt 3 \sin x{(\cos x)^{\frac{1}{2}}}\) for \(0 \le x \le \frac{\pi }{2}\) . Find the volume generated when the curve of g is revolved through \(2\pi \) about the x-axis.

[7]
c.
Answer/Explanation

Markscheme

(i) range of f is \([ – 1{\text{, }}1]\) , \(( – 1 \le f(x) \le 1)\)     A2     N2

(ii) \({\sin ^3}x \Rightarrow 1 \Rightarrow \sin x = 1\)     A1

justification for one solution on \([0{\text{, }}2\pi ]\)    R1

e.g. \(x = \frac{\pi }{2}\) , unit circle, sketch of \(\sin x\)

1 solution (seen anywhere)     A1     N1

[5 marks]

a.

\(f'(x) = 3{\sin ^2}x\cos x\)     A2     N2

[2 marks]

b.

using \(V = \int_a^b {\pi {y^2}{\rm{d}}x} \)     (M1)

\(V = \int_0^{\frac{\pi }{2}} {\pi (\sqrt 3 } \sin x{\cos ^{\frac{1}{2}}}x{)^2}{\rm{d}}x\)     (A1)

\( = \pi \int_0^{\frac{\pi }{2}} {3{{\sin }^2}x\cos x{\rm{d}}x} \)     A1

\(V = \pi \left[ {{{\sin }^3}x} \right]_0^{\frac{\pi }{2}}\) \(\left( { = \pi \left( {{{\sin }^3}\left( {\frac{\pi }{2}} \right) – {{\sin }^3}0} \right)} \right)\)     A2

evidence of using \(\sin \frac{\pi }{2} = 1\) and \(\sin 0 = 0\)     (A1)

e.g. \(\pi \left( {1 – 0} \right)\)

\(V = \pi \)     A1     N1

[7 marks]

c.

Question

Let \(f(x) = \frac{{2x}}{{{x^2} + 5}}\).

Use the quotient rule to show that \(f'(x) = \frac{{10 – 2{x^2}}}{{{{({x^2} + 5)}^2}}}\).

[4]
a.

Find \(\int {\frac{{2x}}{{{x^2} + 5}}{\text{d}}x} \).

[4]
b.

The following diagram shows part of the graph of \(f\).

The shaded region is enclosed by the graph of \(f\), the \(x\)-axis, and the lines \(x = \sqrt 5 \) and \(x = q\). This region has an area of \(\ln 7\). Find the value of \(q\).

[7]
c.
Answer/Explanation

Markscheme

derivative of \(2x\) is \(2\) (must be seen in quotient rule)     (A1)

derivative of \({x^2} + 5\) is \(2x\) (must be seen in quotient rule)     (A1)

correct substitution into quotient rule     A1

eg     \(\frac{{({x^2} + 5)(2) – (2x)(2x)}}{{{{({x^2} + 5)}^2}}},{\text{ }}\frac{{2({x^2} + 5) – 4{x^2}}}{{{{({x^2} + 5)}^2}}}\)

correct working which clearly leads to given answer   A1

eg    \(\frac{{2{x^2} + 10 – 4{x^2}}}{{{{({x^2} + 5)}^2}}},{\text{ }}\frac{{2{x^2} + 10 – 4{x^2}}}{{{x^4} + 10{x^2} + 25}}\)

\(f'(x) = \frac{{10 – 2{x^2}}}{{{{({x^2} + 5)}^2}}}\)     AG     N0

[4 marks]

a.

valid approach using substitution or inspection     (M1)

eg     \(u = {x^2} + 5,{\text{ d}}u = 2x{\text{d}}x,{\text{ }}\frac{1}{2}\ln ({x^2} + 5)\)

\(\int {\frac{{2x}}{{{x^2} + 5}}{\text{d}}x = \int {\frac{1}{u}{\text{d}}u} } \)     (A1)

\(\int {\frac{1}{u}{\text{d}}u = \ln u + c} \)     (A1)

\(\ln ({x^2} + 5) + c\)     A1     N4

[4 marks]

b.

correct expression for area     (A1)

eg     \(\left[ {\ln \left( {{x^2} + 5} \right)} \right]_{\sqrt 5 }^q,{\text{ }}\int\limits_{\sqrt 5 }^q {\frac{{2x}}{{{x^2} + 5}}{\text{d}}x} \)

substituting limits into their integrated function and subtracting (in either order)     (M1)

eg     \(\ln ({q^2} + 5) – \ln \left( {{{\sqrt 5 }^2} + 5} \right)\)

correct working     (A1)

eg     \(\ln \left( {{q^2} + 5} \right) – \ln 10,{\text{ }}\ln \frac{{{q^2} + 5}}{{10}}\)

equating their expression to \(\ln 7\) (seen anywhere)     (M1)

eg     \(\ln \left( {{q^2} + 5} \right) – \ln 10 = \ln 7,{\text{ }}\ln \frac{{{q^2} + 5}}{{10}} = \ln 7,{\text{ }}\ln ({q^2} + 5) = \ln 7 + \ln 10\)

correct equation without logs     (A1)

eg     \(\frac{{{q^2} + 5}}{{10}} = 7,{\text{ }}{q^2} + 5 = 70\)

\({q^2} = 65\)     (A1)

\(q = \sqrt {65} \)     A1     N3

 

Note: Award A0 for \(q =  \pm \sqrt {65} \).

 

[7 marks]

c.

Question

The following diagram shows the graph of \(f(x) = \frac{x}{{{x^2} + 1}}\), for \(0 \le x \le 4\), and the line \(x = 4\).

Let \(R\) be the region enclosed by the graph of \(f\) , the \(x\)-axis and the line \(x = 4\).

Find the area of \(R\).

Answer/Explanation

Markscheme

substitution of limits or function     (A1)

eg\(\;\;\;A = \int_0^4 {f(x),{\text{ }}\int {\frac{x}{{{x^2} + 1}}{\text{d}}x} } \)

correct integration by substitution/inspection     A2

\(\frac{1}{2}\ln ({x^2} + 1)\)

substituting limits into their integrated function and subtracting (in any order)     (M1)

eg\(\;\;\;\frac{1}{2}\left( {\ln ({4^2} + 1) – \ln ({0^2} + 1)} \right)\)

correct working     A1

eg\(\;\;\;\frac{1}{2}\left( {\ln ({4^2} + 1) – \ln ({0^2} + 1)} \right),{\text{ }}\frac{1}{2}\left( {\ln (17) – \ln (1)} \right),{\text{ }}\frac{1}{2}\ln 17 – 0\)

\(A = \frac{1}{2}\ln (17)\)     A1     N3

Note:     Exception to FT rule. Allow full FT on incorrect integration involving a \(\ln \) function.

[6 marks]

Question

Let \(f'(x) = \frac{{6 – 2x}}{{6x – {x^2}}}\), for \(0 < x < 6\).

The graph of \(f\) has a maximum point at P.

The \(y\)-coordinate of P is \(\ln 27\).

Find the \(x\)-coordinate of P.

[3]
a.

Find \(f(x)\), expressing your answer as a single logarithm.

[8]
b.

The graph of \(f\) is transformed by a vertical stretch with scale factor \(\frac{1}{{\ln 3}}\). The image of P under this transformation has coordinates \((a,{\text{ }}b)\).

Find the value of \(a\) and of \(b\), where \(a,{\text{ }}b \in \mathbb{N}\).

[[N/A]]
c.
Answer/Explanation

Markscheme

recognizing \(f'(x) = 0\)     (M1)

correct working     (A1)

eg\(\,\,\,\,\,\)\(6 – 2x = 0\)

\(x = 3\)    A1     N2

[3 marks]

a.

evidence of integration     (M1)

eg\(\,\,\,\,\,\)\(\int {f’,{\text{ }}\int {\frac{{6 – 2x}}{{6x – {x^2}}}{\text{d}}x} } \)

using substitution     (A1)

eg\(\,\,\,\,\,\)\(\int {\frac{1}{u}{\text{d}}u} \) where \(u = 6x – {x^2}\)

correct integral     A1

eg\(\,\,\,\,\,\)\(\ln (u) + c,{\text{ }}\ln (6x – {x^2})\)

substituting \((3,{\text{ }}\ln 27)\) into their integrated expression (must have \(c\))     (M1)

eg\(\,\,\,\,\,\)\(\ln (6 \times 3 – {3^2}) + c = \ln 27,{\text{ }}\ln (18 – 9) + \ln k = \ln 27\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(c = \ln 27 – \ln 9\)

EITHER

\(c = \ln 3\)    (A1)

attempt to substitute their value of \(c\) into \(f(x)\)     (M1)

eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln 3\)     A1     N4

OR

attempt to substitute their value of \(c\) into \(f(x)\)     (M1)

eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln 27 – \ln 9\)

correct use of a log law     (A1)

eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln \left( {\frac{{27}}{9}} \right),{\text{ }}f(x) = \ln \left( {27(6x – {x^2})} \right) – \ln 9\)

\(f(x) = \ln \left( {3(6x – {x^2})} \right)\)    A1     N4

[8 marks]

b.

\(a = 3\)    A1     N1

correct working     A1

eg\(\,\,\,\,\,\)\(\frac{{\ln 27}}{{\ln 3}}\)

correct use of log law     (A1)

eg\(\,\,\,\,\,\)\(\frac{{3\ln 3}}{{\ln 3}},{\text{ }}{\log _3}27\)

\(b = 3\)    A1     N2

[4 marks]

c.

Question

Find \(\int {x{{\text{e}}^{{x^2} – 1}}{\text{d}}x} \).

[4]
a.

Find \(f(x)\), given that \(f’(x) = x{{\text{e}}^{{x^2} – 1}}\) and \(f( – 1) = 3\).

[3]
b.
Answer/Explanation

Markscheme

valid approach to set up integration by substitution/inspection     (M1)

eg\(\,\,\,\,\,\)\(u = {x^2} – 1,{\text{ d}}u = 2x,{\text{ }}\int {2x{{\text{e}}^{{x^2} – 1}}{\text{d}}x} \)

correct expression     (A1)

eg\(\,\,\,\,\,\)\(\frac{1}{2}\int {2x{{\text{e}}^{{x^2} – 1}}{\text{d}}x,{\text{ }}\frac{1}{2}\int {{{\text{e}}^u}{\text{d}}u} } \)

\(\frac{1}{2}{{\text{e}}^{{x^2} – 1}} + c\)     A2     N4

Notes: Award A1 if missing “\( + c\)”.

[4 marks]

a.

substituting \(x =  – 1\) into their answer from (a)     (M1)

eg\(\,\,\,\,\,\)\(\frac{1}{2}{{\text{e}}^0},{\text{ }}\frac{1}{2}{{\text{e}}^{1 – 1}} = 3\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(\frac{1}{2} + c = 3,{\text{ }}c = 2.5\)

\(f(x) = \frac{1}{2}{{\text{e}}^{{x^2} – 1}} + 2.5\)     A1     N2

[3 marks]

b.

Question

Let \(f’(x) = \frac{{3{x^2}}}{{{{({x^3} + 1)}^5}}}\). Given that \(f(0) = 1\), find \(f(x)\).

Answer/Explanation

Markscheme

valid approach     (M1)

eg\(\,\,\,\,\,\)\(\int {f'{\text{d}}x,{\text{ }}\int {\frac{{3{x^2}}}{{{{({x^3} + 1)}^5}}}{\text{d}}x} } \)

correct integration by substitution/inspection     A2

eg\(\,\,\,\,\,\)\(f(x) =  – \frac{1}{4}{({x^3} + 1)^{ – 4}} + c,{\text{ }}\frac{{ – 1}}{{4{{({x^3} + 1)}^4}}}\)

correct substitution into their integrated function (must include \(c\))     M1

eg\(\,\,\,\,\,\)\(1 = \frac{{ – 1}}{{4{{({0^3} + 1)}^4}}} + c,{\text{ }} – \frac{1}{4} + c = 1\)

Note:     Award M0 if candidates substitute into \(f’\) or \(f’’\).

\(c = \frac{5}{4}\)     (A1)

\(f(x) =  – \frac{1}{4}{({x^3} + 1)^{ – 4}} + \frac{5}{4}{\text{ }}\left( { = \frac{{ – 1}}{{4{{({x^3} + 1)}^4}}} + \frac{5}{4},{\text{ }}\frac{{5{{({x^3} + 1)}^4} – 1}}{{4{{({x^3} + 1)}^4}}}} \right)\)     A1     N4

[6 marks]

Question

Let \(f\left( x \right) = \frac{1}{{\sqrt {2x – 1} }}\), for \(x > \frac{1}{2}\).

Find \(\int {{{\left( {f\left( x \right)} \right)}^2}{\text{d}}x} \).

[3]
a.

Part of the graph of f is shown in the following diagram.

The shaded region R is enclosed by the graph of f, the x-axis, and the lines x = 1 and x = 9 . Find the volume of the solid formed when R is revolved 360° about the x-axis.

[4]
b.
Answer/Explanation

Markscheme

correct working      (A1)

eg   \(\int {\frac{1}{{2x – 1}}{\text{d}}x,\,\,\int {{{\left( {2x – 1} \right)}^{ – 1}},\,\,\frac{1}{{2x – 1}},\,\,\int {{{\left( {\frac{1}{{\sqrt u }}} \right)}^2}\frac{{{\text{d}}u}}{2}} } } \)

\({\int {\left( {f\left( x \right)} \right)} ^2}{\text{d}}x = \frac{1}{2}{\text{ln}}\left( {2x – 1} \right) + c\)      A2 N3

Note: Award A1 for \(\frac{1}{2}{\text{ln}}\left( {2x – 1} \right)\).

[3 marks]

a.

attempt to substitute either limits or the function into formula involving f 2 (accept absence of \(\pi \) / dx)     (M1)

eg   \(\int_1^9 {{y^2}{\text{d}}x,\,\,} \pi {\int {\left( {\frac{1}{{\sqrt {2x – 1} }}} \right)} ^2}{\text{d}}x,\,\,\left[ {\frac{1}{2}{\text{ln}}\left( {2x – 1} \right)} \right]_1^9\)

substituting limits into their integral and subtracting (in any order)     (M1)

eg  \(\frac{\pi }{2}\left( {{\text{ln}}\left( {17} \right) – {\text{ln}}\left( 1 \right)} \right),\,\,\pi \left( {0 – \frac{1}{2}{\text{ln}}\left( {2 \times 9 – 1} \right)} \right)\)

correct working involving calculating a log value or using log law     (A1)

eg  \({\text{ln}}\left( 1 \right) = 0,\,\,{\text{ln}}\left( {\frac{{17}}{1}} \right)\)

\(\frac{\pi }{2}{\text{ln}}17\,\,\,\,\left( {{\text{accept }}\pi {\text{ln}}\sqrt {17} } \right)\)    A1 N3

Note: Full FT may be awarded as normal, from their incorrect answer in part (a), however, do not award the final two A marks unless they involve logarithms.

[4 marks]

b.

Question

Find \(\int {\frac{{{{\rm{e}}^x}}}{{1 + {{\rm{e}}^x}}}} {\rm{d}}x\) .

[3]
a.

Find \(\int {\sin 3x\cos 3x{\rm{d}}x} \) .

[4]
b.
Answer/Explanation

Markscheme

attempt to use substitution or inspection     M1

e.g. \(u = 1 + {{\rm{e}}^x}\) so \(\frac{{{\rm{d}}u}}{{{\rm{d}}x}} = {{\rm{e}}^x}\)

correct working     A1

e.g. \(\int {\frac{{{\rm{d}}u}}{u}}  = \ln u\)

\(\ln (1 + {{\rm{e}}^x}) + C\)     A1     N3

[3 marks]

a.

METHOD 1

attempt to use substitution or inspection     M1

e.g. let \(u = \sin 3x\)

\(\frac{{{\rm{d}}u}}{{{\rm{d}}x}} = 3\cos 3x\)     A1

\(\frac{1}{3}\int {u{\rm{d}}u = } \frac{1}{3} \times \frac{{{u^2}}}{2} + C\)    A1

\(\int {\sin 3x\cos 3x{\rm{d}}x = } \frac{{{{\sin }^2}3x}}{6} + C\)     A1     N2

METHOD 2

attempt to use substitution or inspection     M1

e.g. let \(u = \cos 3x\)

\(\frac{{{\rm{d}}u}}{{{\rm{d}}x}} = – 3\sin 3x\)     A1

\( – \frac{1}{3}\int {u{\rm{d}}u = } – \frac{1}{3} \times \frac{{{u^2}}}{2} + C\)     A1

\(\int {\sin 3x\cos 3x{\rm{d}}x = } \frac{{{{\cos }^2}3x}}{6} + C\)     A1     N2

METHOD 3

recognizing double angle     M1

correct working     A1

e.g. \(\frac{1}{2}\sin 6x\)

\(\int {\sin 6x{\rm{d}}x = } \frac{{ – \cos 6x}}{6} + C\)     A1

\(\int {\frac{1}{2}\sin 6x{\rm{d}}x = – \frac{{\cos 6x}}{{12}}}  + C\)     A1     N2

[4 marks]

b.

Leave a Reply