6. (i) Find
$$\int \cot ^ { 2 } 2 x \mathrm {~d} x$$
(ii) Use the substitution \(u ^ { 2 } = x + 1\) to evaluate
$$\int _ { 0 } ^ { 3 } \frac { x ^ { 2 } } { \sqrt { x + 1 } } \mathrm {~d} x$$
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Question 6:
Part (i):
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(= \int (\cosec^2 2x - 1) \, dx\) M1
\(= -\frac{1}{2}\cot 2x - x + c\) M1 A1
Part (ii):
Answer Marks
Guidance
Answer/Working Mark
Guidance
\(u^2 = x + 1 \Rightarrow x = u^2 - 1\), \(\frac{dx}{du} = 2u\) 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
\(= \left(\frac{64}{5} - \frac{32}{3} + 4\right) - \left(\frac{2}{5} - \frac{4}{3} + 2\right) = 5\frac{1}{15}\) M1 A1
(10)
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# Question 6:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $= \int (\cosec^2 2x - 1) \, dx$ | M1 | |
| $= -\frac{1}{2}\cot 2x - x + c$ | M1 A1 | |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u^2 = x + 1 \Rightarrow x = u^2 - 1$, $\frac{dx}{du} = 2u$ | 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 | |
| $= \left(\frac{64}{5} - \frac{32}{3} + 4\right) - \left(\frac{2}{5} - \frac{4}{3} + 2\right) = 5\frac{1}{15}$ | M1 A1 | **(10)** |
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6. (i) Find
$$\int \cot ^ { 2 } 2 x \mathrm {~d} x$$
(ii) Use the substitution $u ^ { 2 } = x + 1$ to evaluate
$$\int _ { 0 } ^ { 3 } \frac { x ^ { 2 } } { \sqrt { x + 1 } } \mathrm {~d} x$$
\hfill \mbox{\textit{OCR C4 Q6 [10]}}