10 It is given that \(I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \frac { \sin ^ { 2 n } x } { \cos x } \mathrm {~d} x\), where \(n \geqslant 0\). Show that
$$I _ { n } - I _ { n + 1 } = \frac { 2 ^ { - \left( n + \frac { 1 } { 2 } \right) } } { 2 n + 1 }$$
Hence show that \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \frac { \sin ^ { 6 } x } { \cos x } \mathrm {~d} x = \ln ( 1 + \sqrt { } 2 ) - \frac { 73 } { 120 } \sqrt { } 2\).
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Question 10:
Answer Marks
Guidance
Answer/Working Mark
Notes
\(I_{n+1} = \int_0^{\frac{1}{4}\pi}\frac{\sin^{2n+2}x}{\cos x}\,dx\) B1
LR
\(I_{n+1} = \int_0^{\frac{1}{4}\pi}\frac{(1-\cos^2 x)\sin^{2n}x}{\cos x}\,dx\) M1
LR
\(= I_n - \int_0^{\frac{1}{4}\pi}\cos x\sin^{2n}x\,dx\) A1
\(= I_n - \left[\frac{\sin^{2n+1}x}{2n+1}\right]_0^{\frac{1}{4}\pi}\) M1
\(\Rightarrow I_n - I_{n+1} = \frac{2^{-(n+\frac{1}{2})}}{2n+1}\) A1 [5]
AG
\(I_0 = \int_0^{\frac{1}{4}\pi}\sec x\,dx = \left[\ln \sec x + \tan x
\right]_0^{\frac{1}{4}\pi} = \ln(1+\sqrt{2})\)
\(I_1 = \ln(1+\sqrt{2}) - \frac{1}{\sqrt{2}}\) M1
\(I_2 = \ln(1+\sqrt{2}) - \frac{1}{\sqrt{2}} - \frac{1}{6\sqrt{2}}\) A1
\(I_3 = \ln(1+\sqrt{2}) - \frac{1}{\sqrt{2}} - \frac{1}{6\sqrt{2}} - \frac{1}{20\sqrt{2}} = \ln(1+\sqrt{2}) - \frac{73\sqrt{2}}{120}\) A1 [5]
AG
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# Question 10:
| Answer/Working | Mark | Notes |
|---|---|---|
| $I_{n+1} = \int_0^{\frac{1}{4}\pi}\frac{\sin^{2n+2}x}{\cos x}\,dx$ | B1 | LR |
| $I_{n+1} = \int_0^{\frac{1}{4}\pi}\frac{(1-\cos^2 x)\sin^{2n}x}{\cos x}\,dx$ | M1 | LR |
| $= I_n - \int_0^{\frac{1}{4}\pi}\cos x\sin^{2n}x\,dx$ | A1 | |
| $= I_n - \left[\frac{\sin^{2n+1}x}{2n+1}\right]_0^{\frac{1}{4}\pi}$ | M1 | |
| $\Rightarrow I_n - I_{n+1} = \frac{2^{-(n+\frac{1}{2})}}{2n+1}$ | A1 [5] | AG |
| $I_0 = \int_0^{\frac{1}{4}\pi}\sec x\,dx = \left[\ln|\sec x + \tan x|\right]_0^{\frac{1}{4}\pi} = \ln(1+\sqrt{2})$ | M1A1 | |
| $I_1 = \ln(1+\sqrt{2}) - \frac{1}{\sqrt{2}}$ | M1 | |
| $I_2 = \ln(1+\sqrt{2}) - \frac{1}{\sqrt{2}} - \frac{1}{6\sqrt{2}}$ | A1 | |
| $I_3 = \ln(1+\sqrt{2}) - \frac{1}{\sqrt{2}} - \frac{1}{6\sqrt{2}} - \frac{1}{20\sqrt{2}} = \ln(1+\sqrt{2}) - \frac{73\sqrt{2}}{120}$ | A1 [5] | AG |
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10 It is given that $I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \frac { \sin ^ { 2 n } x } { \cos x } \mathrm {~d} x$, where $n \geqslant 0$. Show that
$$I _ { n } - I _ { n + 1 } = \frac { 2 ^ { - \left( n + \frac { 1 } { 2 } \right) } } { 2 n + 1 }$$
Hence show that $\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \frac { \sin ^ { 6 } x } { \cos x } \mathrm {~d} x = \ln ( 1 + \sqrt { } 2 ) - \frac { 73 } { 120 } \sqrt { } 2$.
\hfill \mbox{\textit{CAIE FP1 2014 Q10 [10]}}