# 2018 Nov Paper 1 - Q10

For simplicity for IBP we will use the notation:

\displaystyle \int uv' = uv - \int u'v

and we will talk about u and v throughout.

(a)

Doing IBP with u = e^x and v' = \cos 2x, we have u' = e^x and v = \dfrac 1 2 \sin 2x, and so:

\displaystyle \int e^x \cos 2x \mathrm dx = \frac 1 2 e^x \sin 2 x - \frac 1 2 \int e^x \sin 2x \mathrm dx

Similarly, IBP with u = e^x and v' = \sin 2x gives u' = e^x and v = -\dfrac 1 2 \cos 2x, so that:

\displaystyle \int e^x \sin 2x \mathrm dx = -\frac 1 2 e^x \cos 2x + \frac 1 2 \int e^x \cos 2x \mathrm dx

substituting this into the above gives:

\begin{align*}\int e^x \cos 2x \mathrm dx &= \frac 1 2 e^x \sin 2x - \frac 1 2 \left(-\frac 1 2 e^x \cos 2x + \frac 1 2 \int e^x \cos 2x \mathrm dx\right) + C \\ & = \frac 1 2 e^x \sin 2x + \frac 1 4 e^x \cos 2x - \frac 1 4 \int e^x \cos 2x \mathrm dx + C\end{align*}

So:

\displaystyle \frac 5 4 \int e^x \cos 2x \mathrm dx = \frac 1 2 e^x \sin 2x + \frac 1 4 e^x \cos 2x + C

multiplying by \dfrac 4 5 gives:

\displaystyle \int e^x \cos 2x \mathrm dx = \frac 2 5 e^x \sin 2x + \frac 1 5 e^x \cos 2x + c

(b)

We have, using the identity \displaystyle \cos^2 x = \frac {1 + \cos 2x} 2 and part (a):

\begin{align*}\int e^x \cos^2 x \mathrm dx & = \frac 1 2 \int e^x (1 + \cos 2x) \mathrm dx \\ & = \frac 1 2 \int e^x \mathrm dx + \frac 1 2 \int e^x \cos 2x \mathrm dx \\ & = \frac 1 2 e^x + \frac 1 2 \left(\frac 2 5 e^x \sin 2x + \frac 1 5 e^x \cos 2x\right) + c \\ & = \frac 1 2 e^x + \frac 1 5 e^x \sin 2x + \frac 1 {10} e^x \cos 2x + c\end{align*}

We have using the product rule then chain rule:

\begin{align*}f'(x) & = e^x \cos^2 x - 2 e^x \sin x \cos x \\ & = e^x \cos x \left(\cos x - 2 \sin x\right)\end{align*}

So f'(x) = 0 either when \cos x = 0 (corresponding to points like B and D since at these points, we have f(x) = 0) or \cos x - 2 \sin x = 0. (of which A and C are the first two points for x > 0) The latter case gives \tan x = \dfrac 1 2, so x = n \pi + \arctan \dfrac 1 2 for some integer n.

So the x-coordinate of A is \arctan \dfrac 1 2 and the x-coordinate of C is \displaystyle \pi + \arctan \frac 1 2.

(d)

At B and D we have e^x \cos2 x = 0. Since e^x is non-zero, we have \cos x = 0, this has solutions x = \dfrac \pi 2, \dfrac {3 \pi} 2, \ldots. So B has x-coordinate \dfrac \pi 2 and D has x-coordinate \dfrac {3 \pi} 2. So the desired area is:

\begin{align*}\int_{\frac \pi 2}^{\frac {3 \pi} 2} e^x \cos^2 x \mathrm dx & = \left[\frac {e^x} 5 \sin 2x + \frac {e^x} {10} \cos 2x + \frac {e^x} 2 + C\right]_{\frac \pi 2}^{\frac {3 \pi} 2} \\ & = \frac {e^{\frac {3 \pi} 2}} 5 \sin 3 \pi + \frac {e^{\frac {3 \pi} 2}} {10} \cos 3 \pi + \frac {e^{\frac {3 \pi} 2}} 2 \\ & -\frac {e^{\frac \pi 2}} 5 \sin \pi - \frac {e^{\frac \pi 2}} {10} \cos \pi - \frac {e^{\frac \pi 2}} 2 \\ & = -\frac 1 {10} e^{\frac {3 \pi} 2} + \frac 1 {10} e^{\frac \pi 2} + \frac {e^{\frac {3 \pi} 2}} 2 - \frac {e^{\frac \pi 2}} 2 \\ & = \frac 2 5\left( e^{\frac {3 \pi} 2} - e^{\frac \pi 2}\right)\end{align*}

2 Likes

Do you also do the IB?

Nope, this is the first IB maths question I’ve done!

Oh this is amazing