9 Using the substitution \(u = \cos \theta\), or any other method, find \(\int \sin \theta \cos ^ { 2 } \theta d \theta\).
It is given that \(I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \sin ^ { n } \theta \cos ^ { 2 } \theta \mathrm {~d} \theta\), for \(n \geqslant 0\). Show that, for \(n \geqslant 2\),
$$I _ { n } = \frac { n - 1 } { n + 2 } I _ { n - 2 }$$
Hence find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \sin ^ { 4 } \theta \cos ^ { 2 } \theta d \theta\).
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Question 9:
Answer Marks
Guidance
Working/Answer Mark
Guidance
\(\int\sin\theta\cos^2\theta\,d\theta = -\frac{1}{3}\cos^3\theta + c\) B1
Ignore omission of \(c\)
\(I_n = \int_0^{\frac{\pi}{2}}\sin^n\theta\cos^2\theta\,d\theta = \left[-\sin^{n-1}\theta\cdot\frac{\cos^3\theta}{3}\right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}}(n-1)\sin^{n-2}\theta\cos\theta\cdot\frac{\cos^3\theta}{3}\,d\theta\) M1A1
\(= 0 + \int_0^{\frac{\pi}{2}}\frac{(n-1)}{3}\sin^{n-2}\theta\cos^2\theta(1-\sin^2\theta)\,d\theta\) M1A1
N.B. Limits not required for both M marks; parts can use \(u=\cos x\), \(\frac{dv}{dx}=\sin^n x\cos x\)
\(= \frac{(n-1)}{3}(I_{n-2} - I_n)\) \(\Rightarrow I_n = \frac{(n-1)}{(n+2)}I_{n-2}\) A1
AG
Total [5]
\(I_0 = \int_0^{\frac{\pi}{2}}\cos^2\theta\,d\theta = \frac{1}{2}\int_0^{\frac{\pi}{2}}(\cos 2\theta+1)\,d\theta = \frac{1}{2}\left[\frac{\sin 2\theta}{2}+\theta\right]_0^{\frac{\pi}{2}} = \frac{\pi}{4}\) M1A1
Or \(I_2\)
\(\therefore I_4 = \frac{3}{6}\times\frac{1}{4}\times\frac{\pi}{4} = \frac{\pi}{32}\) M1A1
CAO
Total [4]
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# Question 9:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int\sin\theta\cos^2\theta\,d\theta = -\frac{1}{3}\cos^3\theta + c$ | B1 | Ignore omission of $c$ |
| $I_n = \int_0^{\frac{\pi}{2}}\sin^n\theta\cos^2\theta\,d\theta = \left[-\sin^{n-1}\theta\cdot\frac{\cos^3\theta}{3}\right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}}(n-1)\sin^{n-2}\theta\cos\theta\cdot\frac{\cos^3\theta}{3}\,d\theta$ | M1A1 | |
| $= 0 + \int_0^{\frac{\pi}{2}}\frac{(n-1)}{3}\sin^{n-2}\theta\cos^2\theta(1-\sin^2\theta)\,d\theta$ | M1A1 | N.B. Limits not required for both M marks; parts can use $u=\cos x$, $\frac{dv}{dx}=\sin^n x\cos x$ |
| $= \frac{(n-1)}{3}(I_{n-2} - I_n)$ | | |
| $\Rightarrow I_n = \frac{(n-1)}{(n+2)}I_{n-2}$ | A1 | AG |
| Total | [5] | |
| $I_0 = \int_0^{\frac{\pi}{2}}\cos^2\theta\,d\theta = \frac{1}{2}\int_0^{\frac{\pi}{2}}(\cos 2\theta+1)\,d\theta = \frac{1}{2}\left[\frac{\sin 2\theta}{2}+\theta\right]_0^{\frac{\pi}{2}} = \frac{\pi}{4}$ | M1A1 | Or $I_2$ |
| $\therefore I_4 = \frac{3}{6}\times\frac{1}{4}\times\frac{\pi}{4} = \frac{\pi}{32}$ | M1A1 | CAO |
| Total | [4] | |
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9 Using the substitution $u = \cos \theta$, or any other method, find $\int \sin \theta \cos ^ { 2 } \theta d \theta$.
It is given that $I _ { n } = \int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \sin ^ { n } \theta \cos ^ { 2 } \theta \mathrm {~d} \theta$, for $n \geqslant 0$. Show that, for $n \geqslant 2$,
$$I _ { n } = \frac { n - 1 } { n + 2 } I _ { n - 2 }$$
Hence find the exact value of $\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \sin ^ { 4 } \theta \cos ^ { 2 } \theta d \theta$.
\hfill \mbox{\textit{CAIE FP1 2014 Q9 [10]}}