Standard +0.3 Part (a) requires the standard identity tan²x = sec²x - 1, making it straightforward. Part (b) is a routine integration by parts with ln x. Part (c) involves a guided substitution with algebraic manipulation but follows a clear path once u is substituted. All three parts are standard C4 techniques with no novel problem-solving required, making this slightly easier than average.
6. (a) Find \(\int \tan ^ { 2 } x \mathrm {~d} x\).
(b) Use integration by parts to find \(\int \frac { 1 } { x ^ { 3 } } \ln x \mathrm {~d} x\).
(c) Use the substitution \(u = 1 + e ^ { x }\) to show that
$$\int \frac { \mathrm { e } ^ { 3 x } } { 1 + \mathrm { e } ^ { x } } \mathrm {~d} x = \frac { 1 } { 2 } \mathrm { e } ^ { 2 x } - \mathrm { e } ^ { x } + \ln \left( 1 + \mathrm { e } ^ { x } \right) + k$$
where \(k\) is a constant.
\(\int\tan^2 x\, dx\); use \(\tan^2 A=\sec^2 A-1\)
M1 oe
The correct underlined identity
\(=\int\sec^2 x-1\, dx = \tan x - x\,(+c)\)
A1
Correct integration with/without \(+c\)
[2]
Part (b):
Answer
Marks
Guidance
Answer/Working
Mark
Guidance
\(\int\frac{1}{x^3}\ln x\, dx\); let \(u=\ln x \Rightarrow \frac{du}{dx}=\frac{1}{x}\); \(\frac{dv}{dx}=x^{-3}\Rightarrow v=\frac{x^{-2}}{-2}=\frac{-1}{2x^2}\)
Differentiating to find any one of the three underlined expressions
\(=\int\frac{e^{2x}\cdot e^x}{1+e^x}\,dx=\int\frac{(u-1)^2\cdot e^x}{u}\cdot\frac{1}{e^x}\,du\) or \(\int\frac{(u-1)^3}{u}\cdot\frac{1}{u-1}\,du\)
M1*
Attempt to substitute \(e^{2x}=f(u)\), their \(\frac{dx}{du}=\frac{1}{e^x}\) and \(u=1+e^x\), or \(e^{3x}=f(u)\), their \(\frac{dx}{du}=\frac{1}{u-1}\) and \(u=1+e^x\)
An attempt to multiply out numerator to give at least three terms and divide through each term by \(u\)
\(=\frac{u^2}{2}-2u+\ln u\,(+c)\)
A1
Correct integration with/without \(+c\)
\(=\frac{(1+e^x)^2}{2}-2(1+e^x)+\ln(1+e^x)+c\)
dM1*
Substitutes \(u=1+e^x\) back into integrated expression with at least two terms
\(=\frac{1}{2}e^{2x}-e^x+\ln(1+e^x)+k\) AG
A1 cso
\(\frac{1}{2}e^{2x}-e^x+\ln(1+e^x)+k\); must use \(a+c\) and \("-\frac{3}{2}"\) combined
[7]
13 marks
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\tan^2 x\, dx$; use $\tan^2 A=\sec^2 A-1$ | M1 oe | The correct underlined identity |
| $=\int\sec^2 x-1\, dx = \tan x - x\,(+c)$ | A1 | Correct integration with/without $+c$ |
| **[2]** | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\frac{1}{x^3}\ln x\, dx$; let $u=\ln x \Rightarrow \frac{du}{dx}=\frac{1}{x}$; $\frac{dv}{dx}=x^{-3}\Rightarrow v=\frac{x^{-2}}{-2}=\frac{-1}{2x^2}$ | | |
| $=-\frac{1}{2x^2}\ln x-\int-\frac{1}{2x^2}\cdot\frac{1}{x}\,dx$ | M1, A1 | Use of integration by parts formula in the correct direction. Correct expression. |
| $=-\frac{1}{2x^2}\ln x+\frac{1}{2}\int\frac{1}{x^3}\,dx$ | M1 | An attempt to multiply through $\frac{k}{x^n}$, $n\in\mathbb{Z}$, $n\geq 2$ by $\frac{1}{x}$ and attempt to integrate |
| $=-\frac{1}{2x^2}\ln x+\frac{1}{2}\left(-\frac{1}{2x^2}\right)(+c)$ | A1 oe | Correct solution with/without $+c$ |
| **[4]** | | Correct direction means $u=\ln x$ |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $u=1+e^x \Rightarrow \frac{du}{dx}=e^x$, $\frac{dx}{du}=\frac{1}{e^x}$, $\frac{dx}{du}=\frac{1}{u-1}$ | B1 | Differentiating to find any one of the three underlined expressions |
| $=\int\frac{e^{2x}\cdot e^x}{1+e^x}\,dx=\int\frac{(u-1)^2\cdot e^x}{u}\cdot\frac{1}{e^x}\,du$ or $\int\frac{(u-1)^3}{u}\cdot\frac{1}{u-1}\,du$ | M1* | Attempt to substitute $e^{2x}=f(u)$, their $\frac{dx}{du}=\frac{1}{e^x}$ and $u=1+e^x$, or $e^{3x}=f(u)$, their $\frac{dx}{du}=\frac{1}{u-1}$ and $u=1+e^x$ |
| $=\int\frac{(u-1)^2}{u}\,du$ | A1 | |
| $=\int\frac{u^2-2u+1}{u}\,du=\int u-2+\frac{1}{u}\,du$ | dM1* | An attempt to multiply out numerator to give at least three terms and divide through each term by $u$ |
| $=\frac{u^2}{2}-2u+\ln u\,(+c)$ | A1 | Correct integration with/without $+c$ |
| $=\frac{(1+e^x)^2}{2}-2(1+e^x)+\ln(1+e^x)+c$ | dM1* | Substitutes $u=1+e^x$ back into integrated expression with at least two terms |
| $=\frac{1}{2}e^{2x}-e^x+\ln(1+e^x)+k$ AG | A1 cso | $\frac{1}{2}e^{2x}-e^x+\ln(1+e^x)+k$; must use $a+c$ and $"-\frac{3}{2}"$ combined |
| **[7]** | | |
| **13 marks** | | |
6. (a) Find $\int \tan ^ { 2 } x \mathrm {~d} x$.\\
(b) Use integration by parts to find $\int \frac { 1 } { x ^ { 3 } } \ln x \mathrm {~d} x$.\\
(c) Use the substitution $u = 1 + e ^ { x }$ to show that
$$\int \frac { \mathrm { e } ^ { 3 x } } { 1 + \mathrm { e } ^ { x } } \mathrm {~d} x = \frac { 1 } { 2 } \mathrm { e } ^ { 2 x } - \mathrm { e } ^ { x } + \ln \left( 1 + \mathrm { e } ^ { x } \right) + k$$
where $k$ is a constant.\\
\hfill \mbox{\textit{Edexcel C4 2009 Q6 [13]}}