6. (a) Use the substitution \(x = 2 \sin u\) to evaluate
$$\int _ { 0 } ^ { \sqrt { 3 } } \frac { 1 } { \sqrt { 4 - x ^ { 2 } } } \mathrm {~d} x$$
(b) Use integration by parts to evaluate
$$\int _ { 0 } ^ { \frac { \pi } { 2 } } x \cos x d x$$
6. continued
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(a) \(x = 2\sin u \Rightarrow \frac{dx}{du} = 2\cos u\)M1
\(x = 0 \Rightarrow u = 0, x = \sqrt{3} \Rightarrow u = \frac{\pi}{3}\) B1
\(I = \int_0^{\frac{\pi}{3}}\frac{1}{2\cos u} \times 2\cos u \, du = \int_0^{\frac{\pi}{3}}1 \, du\) A1
\(= [u]_0^{\frac{\pi}{3}} = \frac{\pi}{3} - 0 = \frac{\pi}{3}\) M1 A1
(b) \(u = x, u' = 1, v' = \cos x, v = \sin x\)M1
\(I = [x\sin x]_0^{\frac{\pi}{3}} - \int\sin x \, dx\) A2
\(= [x\sin x + \cos x]_0^{\frac{\pi}{3}}\) M1
\(= (\frac{\pi}{3} + 0) - (0 + 1) = \frac{\pi}{3} - 1\) M1 A1
(11)
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**(a)** $x = 2\sin u \Rightarrow \frac{dx}{du} = 2\cos u$ | M1 |
$x = 0 \Rightarrow u = 0, x = \sqrt{3} \Rightarrow u = \frac{\pi}{3}$ | B1 |
$I = \int_0^{\frac{\pi}{3}}\frac{1}{2\cos u} \times 2\cos u \, du = \int_0^{\frac{\pi}{3}}1 \, du$ | A1 |
$= [u]_0^{\frac{\pi}{3}} = \frac{\pi}{3} - 0 = \frac{\pi}{3}$ | M1 A1 |
**(b)** $u = x, u' = 1, v' = \cos x, v = \sin x$ | M1 |
$I = [x\sin x]_0^{\frac{\pi}{3}} - \int\sin x \, dx$ | A2 |
$= [x\sin x + \cos x]_0^{\frac{\pi}{3}}$ | M1 |
$= (\frac{\pi}{3} + 0) - (0 + 1) = \frac{\pi}{3} - 1$ | M1 A1 | (11)
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6. (a) Use the substitution $x = 2 \sin u$ to evaluate
$$\int _ { 0 } ^ { \sqrt { 3 } } \frac { 1 } { \sqrt { 4 - x ^ { 2 } } } \mathrm {~d} x$$
(b) Use integration by parts to evaluate
$$\int _ { 0 } ^ { \frac { \pi } { 2 } } x \cos x d x$$
6. continued\\
\hfill \mbox{\textit{Edexcel C4 Q6 [11]}}