6. (a) Find
$$\int 2 \sin 3 x \sin 2 x d x$$
(b) Use the substitution \(u ^ { 2 } = x + 1\) to evaluate
$$\int _ { 0 } ^ { 3 } \frac { x ^ { 2 } } { \sqrt { x + 1 } } \mathrm {~d} x$$
6. continued
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(a) \(= \int (\cos x - \cos 5x) dx\)M1 A1
\(= \sin x - \frac{1}{5}\sin 5x + c\) M1 A1
(b) \(u^2 = x + 1 \Rightarrow x = u^2 - 1\), \(\frac{dx}{du} = 2u\)M1
\(x = 0 \Rightarrow u = 1\), \(x = 3 \Rightarrow u = 2\) B1
\(I = \int_1^2 \frac{(u^2-1)^2}{u} \times 2u du = \int_1^2 (2u^4 - 4u^2 + 2) du\) M1 A1
\(= \left[\frac{2}{5}u^5 - \frac{4}{3}u^3 + 2u\right]_1^2\) M1 A1
\(= \left(\frac{64}{5} - \frac{32}{3} + 4\right) - \left(\frac{2}{5} - \frac{4}{3} + 2\right) = 5\frac{1}{15}\) M1 A1
(12 marks)
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**(a)** $= \int (\cos x - \cos 5x) dx$ | M1 A1 |
$= \sin x - \frac{1}{5}\sin 5x + c$ | M1 A1 |
**(b)** $u^2 = x + 1 \Rightarrow x = u^2 - 1$, $\frac{dx}{du} = 2u$ | M1 |
$x = 0 \Rightarrow u = 1$, $x = 3 \Rightarrow u = 2$ | B1 |
$I = \int_1^2 \frac{(u^2-1)^2}{u} \times 2u du = \int_1^2 (2u^4 - 4u^2 + 2) du$ | M1 A1 |
$= \left[\frac{2}{5}u^5 - \frac{4}{3}u^3 + 2u\right]_1^2$ | M1 A1 |
$= \left(\frac{64}{5} - \frac{32}{3} + 4\right) - \left(\frac{2}{5} - \frac{4}{3} + 2\right) = 5\frac{1}{15}$ | M1 A1 | (12 marks)
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6. (a) Find
$$\int 2 \sin 3 x \sin 2 x d x$$
(b) Use the substitution $u ^ { 2 } = x + 1$ to evaluate
$$\int _ { 0 } ^ { 3 } \frac { x ^ { 2 } } { \sqrt { x + 1 } } \mathrm {~d} x$$
6. continued\\
\hfill \mbox{\textit{Edexcel C4 Q6 [12]}}