1 By using the substitution \(t = \tan \frac { 1 } { 2 } \theta\), find \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { 1 } { 1 + \cos \theta } \mathrm { d } \theta\).
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Question 1:
Answer Marks
Guidance
Answer Marks
Guidance
\(\cos\theta = \frac{1-t^2}{1+t^2}\) M1
Using \(t\) substitution for both \(\cos\theta\) and \(d\theta\)
\(\frac{dt}{d\theta} = \frac{1}{2}\sec^2\frac{1}{2}\theta = \frac{1}{2}\left(1+\tan^2\frac{1}{2}\theta\right)\) A1
Subs correct
\(\Rightarrow dt = \frac{1+t^2}{2}\cdot d\theta \Rightarrow d\theta = \frac{2dt}{1+t^2}\) M1
Dealing with limits and attempting integration
\(\Rightarrow I = \int_0^1 \frac{1}{1+\frac{1-t^2}{1+t^2}}\cdot\frac{2dt}{1+t^2} = \int_0^1\frac{1+t^2}{1+t^2+1-t^2}\cdot\frac{2dt}{1+t^2}\) A1
Correct integral
\(\int_0^1\frac{2dt}{2} = [t]_0^1 = 1\) A1
Answer
[5]
Alternative:
Answer Marks
Guidance
Answer Marks
Guidance
\(1+\cos\theta = 2\cos^2\frac{1}{2}\theta\); \(\Rightarrow\int_0^{\frac{\pi}{2}}\frac{1}{1+\cos\theta}d\theta = \frac{1}{2}\int_0^{\frac{\pi}{2}}\frac{1}{\cos^2\frac{1}{2}\theta}d\theta = \frac{1}{2}\int_0^{\frac{\pi}{2}}\sec^2\frac{1}{2}\theta\, d\theta = \frac{1}{2}\left[2\tan\frac{1}{2}\theta\right]_0^{\frac{\pi}{2}} = \tan\frac{\pi}{2}-\tan 0 = 1\) SC3
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# Question 1:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos\theta = \frac{1-t^2}{1+t^2}$ | M1 | Using $t$ substitution for **both** $\cos\theta$ and $d\theta$ |
| $\frac{dt}{d\theta} = \frac{1}{2}\sec^2\frac{1}{2}\theta = \frac{1}{2}\left(1+\tan^2\frac{1}{2}\theta\right)$ | A1 | Subs correct |
| $\Rightarrow dt = \frac{1+t^2}{2}\cdot d\theta \Rightarrow d\theta = \frac{2dt}{1+t^2}$ | M1 | Dealing with limits and attempting integration |
| $\Rightarrow I = \int_0^1 \frac{1}{1+\frac{1-t^2}{1+t^2}}\cdot\frac{2dt}{1+t^2} = \int_0^1\frac{1+t^2}{1+t^2+1-t^2}\cdot\frac{2dt}{1+t^2}$ | A1 | Correct integral |
| $\int_0^1\frac{2dt}{2} = [t]_0^1 = 1$ | A1 | Answer |
| **[5]** | | |
**Alternative:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $1+\cos\theta = 2\cos^2\frac{1}{2}\theta$; $\Rightarrow\int_0^{\frac{\pi}{2}}\frac{1}{1+\cos\theta}d\theta = \frac{1}{2}\int_0^{\frac{\pi}{2}}\frac{1}{\cos^2\frac{1}{2}\theta}d\theta = \frac{1}{2}\int_0^{\frac{\pi}{2}}\sec^2\frac{1}{2}\theta\, d\theta = \frac{1}{2}\left[2\tan\frac{1}{2}\theta\right]_0^{\frac{\pi}{2}} = \tan\frac{\pi}{2}-\tan 0 = 1$ | SC3 | |
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1 By using the substitution $t = \tan \frac { 1 } { 2 } \theta$, find $\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \frac { 1 } { 1 + \cos \theta } \mathrm { d } \theta$.
\hfill \mbox{\textit{OCR FP2 2013 Q1 [5]}}