EXM.3.AHL.TZ0.3

pestleMathematicsAAHLPaper 3EXM· ahl-5-16-integration-by-substitution-parts-and-repeated-partssource ↗

This question will investigate methods for finding definite integrals of powers of trigonometrical functions.

Let ${I_n} = \int\limits_0^{\tfrac{\pi }{2}} {{\text{si}}{{\text{n}}^n}} x\,dx{\text{,}}\,\,n \in \mathbb{N}$ .

Let ${J_n} = \int\limits_0^{\tfrac{\pi }{2}} {{\text{co}}{{\text{s}}^n}} x\,dx{\text{,}}\,\,n \in \mathbb{N}.$

Let ${T_n} = \int\limits_0^{\tfrac{\pi }{4}} {{\text{ta}}{{\text{n}}^n}} x\,dx{\text{,}}\,\,n \in \mathbb{N}$ .

Find the exact values of ${I_0}$ , ${I_1}$ and ${I_2}$ .

[6]

Use integration by parts to show that ${I_n} = \frac{{n - 1}}{n}{I_{n - 2}}{\text{,}}\,\,n \geqslant 2$ .

[5]

b.i.

Explain where the condition $n \geqslant 2$ was used in your proof.

[1]

b.ii.

Hence, find the exact values of ${I_3}$ and ${I_4}$ .

[2]

Use the substitution $x = \frac{\pi }{2} - u$ to show that ${J_n} = {I_n}$ .

[4]

Hence, find the exact values of ${J_{5}}$ and ${J_{6}}$

[2]

Find the exact values of ${T_{0}}$ and ${T_{1}}$ .

[3]

Use the fact that ${\text{ta}}{{\text{n}}^2}x = {\text{se}}{{\text{c}}^2}x - 1$ to show that ${T_n} = \frac{1}{{n - 1}} - {T_{n - 2}}{\text{,}}\,\,n \geqslant 2$ .

[3]

g.i.

Explain where the condition $n \geqslant 2$ was used in your proof.

[1]

g.ii.

Hence, find the exact values of ${T_{2}}$ and ${T_{3}}$ .

[2]

Markscheme / solution

${I_0} = \int\limits_0^{\tfrac{\pi }{2}} 1 \,dx = \left[ x \right]_0^{\tfrac{\pi }{2}} = \frac{\pi }{2}$ M1A1

${I_1} = \int\limits_0^{\tfrac{\pi }{2}} {{\text{sin}}\,x} \,dx = \left[ { - {\text{cos}}\,x} \right]_0^{\tfrac{\pi }{2}} = 1$ M1A1

${I_2} = \int\limits_0^{\tfrac{\pi }{2}} {{\text{si}}{{\text{n}}^2}x} \,dx = \int\limits_0^{\tfrac{\pi }{2}} {\frac{{1 - {\text{cos}}\,2x}}{2}} \,dx = \left[ {\frac{x}{2} - \frac{{{\text{sin}}\,2x}}{4}} \right]_0^{\tfrac{\pi }{2}} = \frac{\pi }{4}$ M1A1

[6 marks]

${u = {\text{si}}{{\text{n}}^{n - 1}}x}$ ${v = - \cos x}$

${\frac{{du}}{{dx}} = \left( {n - 1} \right){\text{si}}{{\text{n}}^{n - 2}}x\,{\text{cos}}\,x}$ ${\frac{{dv}}{{dx}} = {\text{sin}}\,x}$

${I_n} = \left[ { - {\text{si}}{{\text{n}}^{n - 1}}x\,{\text{cos}}\,x} \right]_0^{\tfrac{\pi }{2}} + \int\limits_0^{\tfrac{\pi }{2}} {\left( {n - 1} \right){\text{si}}{{\text{n}}^{n - 2}}x\,{\text{cos}}{\,^2}x} \,dx$ M1A1A1

$= 0 + \int\limits_0^{\tfrac{\pi }{2}} {\left( {n - 1} \right){\text{si}}{{\text{n}}^{n - 2}}x\left( {1 - {\text{si}}{{\text{n}}^2}x} \right)} \,dx = \left( {n - 1} \right)\left( {{I_{n - 2}} - {I_n}} \right)$ M1A1

$\Rightarrow n{I_n} = \left( {n - 1} \right){I_{n - 2}} \Rightarrow {I_n} = \frac{{\left( {n - 1} \right)}}{n}{I_{n - 2}}$ AG

[6 marks]

b.i.

need $n \geqslant 2$ so that ${\text{si}}{{\text{n}}^{n - 1}}\tfrac{\pi }{2} = 0$ in $\left[ { - {\text{si}}{{\text{n}}^{n - 1}}\,x\,{\text{cos}}\,x} \right]_0^{\tfrac{\pi }{2}}$ R1

[1 mark]

b.ii.

${I_3} = \frac{2}{3}{I_1} = \frac{2}{3}\,\,\,\,\,\,\,\,\,\,\,\,\,{I_4} = \frac{3}{4}{I_2} = \frac{{3\pi }}{{16}}$ A1A1

[2 marks]

$x = \frac{\pi }{2} - u \Rightarrow \frac{{dx}}{{du}} = - 1$ A1

${J_n} = \int\limits_0^{\tfrac{\pi }{2}} {{\text{co}}{{\text{s}}^n}} x\,dx\,\, = \int\limits_{\tfrac{\pi }{2}}^0 { - {\text{co}}{{\text{s}}^n}} \left( {\frac{\pi }{2} - u} \right)\,du = - \int\limits_{\tfrac{\pi }{2}}^0 {{\text{si}}{{\text{n}}^n}} u\,du\, = \int\limits_0^{\tfrac{\pi }{2}} {{\text{si}}{{\text{n}}^n}} u\,du\, = {I_n}$ M1A1A1AG

[4 marks]

${J_5} = {I_5} = \frac{4}{5}{I_3} = \frac{4}{5} \times \frac{2}{3} = \frac{8}{{15}}\,\,\,\,\,\,\,\,\,\,{J_6} = {I_6} = \frac{5}{6}{I_4} = \frac{5}{6} \times \frac{{3\pi }}{{16}} = \frac{{5\pi }}{{32}}$ A1A1

[2 marks]

${T_{0\,}}{\text{ = }}\int\limits_0^{\tfrac{\pi }{4}} {1\,dx} = \left[ x \right]_0^{\tfrac{\pi }{4}} = \frac{\pi }{4}$ A1

${T_{1\,}}{\text{ = }}\int\limits_0^{\tfrac{\pi }{4}} {{\text{tan}}\,dx} = \left[ { - \ln \left| {{\text{cos}}\,x} \right|} \right]_0^{\tfrac{\pi }{4}} = - {\text{ln}}\frac{1}{{\sqrt 2 }} = {\text{ln}}\sqrt 2$ M1A1

[3 marks]

${T_n} = \int\limits_0^{\tfrac{\pi }{4}} {{\text{ta}}{{\text{n}}^n}} x\,dx\, = \int\limits_0^{\tfrac{\pi }{4}} {{\text{ta}}{{\text{n}}^{n - 2}}} x\,{\text{ta}}{{\text{n}}^2}x\,dx\, = \,\int\limits_0^{\tfrac{\pi }{4}} {{\text{ta}}{{\text{n}}^{n - 2}}} x\left( {{\text{se}}{{\text{c}}^2}x - 1} \right)\,dx$ M1

$\int\limits_0^{\tfrac{\pi }{4}} {{\text{ta}}{{\text{n}}^{n - 2}}} x\,{\text{se}}{{\text{c}}^2}x\,dx - \int\limits_0^{\tfrac{\pi }{4}} {{\text{ta}}{{\text{n}}^{n - 2}}} x\,dx = \left[ {\frac{{{\text{ta}}{{\text{n}}^{n - 1}}x}}{{n - 1}}} \right]_0^{\tfrac{\pi }{4}} - {T_{n - 2}} = \frac{1}{{n - 1}} - {T_{n - 2}}\,\,\,$ A1A1AG

[3 marks]

g.i.

need $n \geqslant 2$ so that the powers of tan in $\int\limits_0^{\tfrac{\pi }{4}} {{\text{ta}}{{\text{n}}^{n - 2}}} x\,{\text{se}}{{\text{c}}^2}x\,dx - \int\limits_0^{\tfrac{\pi }{4}} {{\text{ta}}{{\text{n}}^{n - 2}}} x\,dx$ are not negative R1

[1 mark]

g.ii.

${T_2} = 1 - {T_0} = 1 - \frac{\pi }{4}$ A1

${T_3} = \frac{1}{2} - {T_1} = \frac{1}{2} - \ln \sqrt 2$ A1

[2 marks]

Examiners’ report

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b.i.

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b.ii.

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g.i.

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g.ii.

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