Advanced Higher Maths
Integration

The key to this integration is in noticing that the numerator is very closely related to the derivative of the denominator.

$$ \small\frac{d}{dx}\normalsize\left(2x^3-2x+1\right)=6x^2-2 $$

So the numerator is exactly half of the derivative of the denominator. Therefore:

$$ \Large\int\normalsize\!\!\small\frac{3x^2\!-\!1}{2x^3\!-\!2x\!+\!1}\normalsize\,dx= \small\frac{1}{2}\normalsize\,ln\vert 2x^3\!-\!2x+1\vert +c $$

This example makes use of one of the standard integrals above:

$$ \Large\int\normalsize\!\!\small\frac{dx}{\sqrt{a^2-x^2}}\normalsize = sin^{-1}\left(\small\frac{x}{a}\normalsize\right)+c $$

\(a\!=\!2\small,\) we have \(\raise 0.2pt{(3x)^2}\) instead of \(\raise 0.2pt{x^2}\) and the entire standard integral is multiplied by \(6\small.\) So:

$$ \begin{eqnarray} \Large\int\normalsize \small\frac{6\,dx}{\sqrt{4-9x^2}} &=& \small\frac{6}{3}\normalsize\,sin^{-1}\small\left(\frac{3x}{2}\right)\normalsize+c \\[6pt] &=& 2\,sin^{-1}\left(\small\frac{3x}{2}\normalsize\right)+c \\[6pt] \end{eqnarray} $$

\(x^2\!-\!2x\!-\!15\) factorises as \((x\!-\!5)(x\!+\!3)\) so:

$$ \frac{3x-7}{(x-5)(x+3)} = \frac{A}{x-5} + \frac{B}{x+3} $$

$$ 3x-7 = A(x+3) + B(x-5) $$

Substituting specific values of \(\raise 0.2pt{x}\) lets us find the values of \(\raise 0.2pt{A}\) and \(\raise 0.2pt{B\small.}\)

\(x\!=\!\!-\!3\implies -16\!=\!B(-8)\implies B\!=\!2\)
\(x\!=\!5\implies 8\!=\!A(8)\implies A\!=\!1\)

So now we are ready to integrate:

$$ \begin{eqnarray} && \large\int\normalsize \small\frac{3x-7}{x^2-2x-15}\normalsize\,dx \\[6pt] &=& \large\int\normalsize \small\left(\frac{1}{x-5}\normalsize+\small\frac{2}{x+3}\right)\normalsize\,dx \\[6pt] &=& ln\vert x-5\vert + 2\,ln\vert x+3\vert+c \\[6pt] &=& ln\vert (x-5)(x+3)^2\vert +c \\[6pt] \end{eqnarray} $$

Note that the last step is a matter of personal preference. The second-last line is a perfectly acceptable final answer.

This example also requires partial fractions, but this time the denominator contains a repeated factor, so the form of the partial fractions is:

$$ \small\frac{A}{x+1}\normalsize+\small\frac{B}{(x+1)^2}\normalsize+\small\frac{C}{2x-1}\normalsize $$

Multiplying through by the denominator, we obtain:

$$ x\!\small+\normalsize\!4\!\small=\normalsize\!A(x\!\small+\normalsize\!1)(2x\!\small-\normalsize\!1)\!\small+\normalsize\!B(2x\!\small-\normalsize\!1)\!\small+\normalsize\!C(x\!\small+\normalsize\!1)^2 $$

Now we can substitute well-chosen values of \(\raise 0.2pt{x}\) to find two of our three unknowns:

$$ x\!=\!\!-\!1\implies 3\!=\!B(-3)\implies B\!=\!\!-\!\!1 $$ $$ x\!=\!\small\frac{1}{2}\normalsize\implies \small\frac{9}{2}\normalsize\!=\!C\small\left(\frac{9}{4}\right)\normalsize\implies C\!=\!2 $$

No substitution will let us find \(A\small.\) Instead, we can equate the constant terms:

So now our integration is fairly easy:

$$ \begin{eqnarray} && \large\int^{\small 2\normalsize}_{\small 1\normalsize}\normalsize \left(-\small\frac{1}{x+1}\normalsize-\small\frac{1}{(x+1)^2}\normalsize+\small\frac{2}{2x-1}\right)\,dx \\[6pt] && \large\int^{\small 2\normalsize}_{\small 1\normalsize}\normalsize \left(-\small\frac{1}{x\!+\!1}\normalsize-{(x\!+\!1)^{-2}}\normalsize+\small\frac{2}{2x\!-\!1}\right)\,dx \\[6pt] &=& \Large[\normalsize-ln\vert x\!+\!1\vert + (x\!+\!1)^{-1} +ln\vert 2x\!-\!1\vert\Large]\normalsize^{\small 2\normalsize}_{\small 1\normalsize}\normalsize \\[6pt] &=& (-ln\,3\!+\!\small\frac{1}{3}\normalsize\!+\!ln\,3)-(-ln\,2\!+\!\small\frac{1}{2}\normalsize\!+\!\cancel{ln\,1}) \\[6pt] &=& \cancel{-ln\,3}\!+\!\small\frac{1}{3}\normalsize\!+\cancel{\!ln\,3}+ln\,2\!-\!\small\frac{1}{2}\normalsize \\[6pt] &=& ln\,2-\small\frac{1}{6} \\[6pt] \end{eqnarray} $$

We start by differentiating \(\raise 0.2pt{u}\) and obtaining an expression for \(\raise 0.2pt{dx\small.}\)

$$ \begin{eqnarray} u &=& tan\,x \\[6pt] \small\frac{du}{dx}\normalsize &=& sec^{2}\,x \\[6pt] du &=& sec^{2}\,x\,dx \\[6pt] dx &=& \small\frac{du}{sec^{2}\,x} \end{eqnarray} $$

Now we subtitute this into the original integration to express it in terms of \(\raise 0.2pt{u\small.}\)

$$ \begin{eqnarray} \large\int\normalsize\!\small\frac{dx}{sin\,x\,cos\,x} &=& \large\int\normalsize\!\small\frac{1}{sin\,x\,cos\,x}\normalsize.\small\frac{du}{sec^{2}\,x}\normalsize \\[6pt] &=& \large\int\normalsize\!\small\frac{cos^{2}\,x}{sin\,x\,cos\,x}\normalsize\,du \\[6pt] &=& \large\int\normalsize\!\small\frac{cos\,x}{sin\,x}\normalsize\,du \\[6pt] &=& \large\int\normalsize\!\small\frac{1}{tan\,x}\normalsize\,du \\[6pt] &=& \large\int\normalsize\!\small\frac{1}{u}\normalsize\,du \\[6pt] &=& ln\vert u\vert+c \\[6pt] &=& ln\vert tan\,x\vert+c \\[6pt] \end{eqnarray} $$

Although integration by parts is another possible method here, we would advise you always to ignore the words "or otherwise".

We start by differentiating \(\raise 0.2pt{u}\small.\)

$$ \begin{eqnarray} u &=& sin\,\theta \\[6pt] \small\frac{du}{d\theta}\normalsize &=& cos\,\theta \\[6pt] \end{eqnarray} $$

Because this is a definite integration, we need to express the upper and lower limits in terms of the new variable.

When \(\theta=\frac{\pi}{2}\small,\) \(u=sin\,\frac{\pi}{2}=1\small.\)
When \(\theta=\frac{\pi}{6}\small,\) \(u=sin\,\frac{\pi}{6}=\frac{1}{2}\small.\)

Now we express the original integration in terms of \(\raise 0.2pt{u\small.}\)

$$ \begin{eqnarray} \int^{\pi/2}_{\pi/6}2\,sin^4\,\theta\,cos\,\theta\,d\theta &=& \int^{1}_{1/2}2\,u^4\,\small\frac{du}{d\theta}\normalsize\,d\theta \\[6pt] &=& 2\,\int^{1}_{1/2}u^4\,du \\[6pt] &=& 2\,\left[\small\frac{1}{5}\normalsize u^5 \right]^1_{1/2} \\[6pt] &=& \small\frac{2}{5}\normalsize \,\Large\left[\normalsize u^5 \Large\right]^{\normalsize1}_{\normalsize1/2} \\[6pt] &=& \small\frac{2}{5}\normalsize\,\left(1^5-\small\left(\frac{1}{2}\right)^{\!5}\normalsize\right) \\[6pt] &=& \small\frac{2}{5}\normalsize\,\left(1-\small\frac{1}{32}\normalsize\right) \\[6pt] &=& \small\frac{2}{5}\normalsize\,\left(\small\frac{31}{32}\normalsize\right) \\[6pt] &=& \small\frac{31}{80}\normalsize \end{eqnarray} $$

This question is a simple 3-mark single use of integration by parts.

First we examine the factors and ask ourselves which we should differentiate and which we should integrate.

As \(ln\,x\) differentiates to \(\large\frac{1}{x}\small,\) this will eliminate the \(ln\,x\) factor, giving us a simple integral.

$$ \begin{matrix} u'=x^4 \:&\: v=ln\,x \\ u=\large\frac{1}{5}\normalsize x^5 \:&\: v'=\large\frac{1}{x} \\ \end{matrix} $$

Now we use integration by parts:

$$ \begin{eqnarray} \int x^4\,ln\,x\,dx &=& uv-\int\,uv'\,dx \\[6pt] &=& \small\frac{1}{5}\normalsize x^5\,ln\,x-\int \small\frac{1}{5}\normalsize x^5.\small\frac{1}{x}\normalsize\,dx \\[6pt] &=& \small\frac{1}{5}\normalsize x^5\,ln\,x-\small\frac{1}{5}\normalsize\int x^4\,dx \\[6pt] &=& \small\frac{1}{5}\normalsize x^5\,ln\,x-\small\frac{1}{5}\normalsize.\small\frac{1}{5}\normalsize x^5+c \\[6pt] &=& \small\frac{1}{5}\normalsize x^5\,ln\,x-\small\frac{1}{25}\normalsize x^5+c \end{eqnarray} $$

This 5-mark question will require us to use integration by parts twice, as you are about to see.

First we examine the factors and ask ourselves which we should differentiate and which we should integrate.

The exponential factor doesn't help us make this decision, because whether we differentiate or integrate it, we will still have some number times \(e^{3x}\small.\)

So we consider the \(x^2\) factor. Clearly differentiating it will reduce the power, so we will differentiate \(x^2\) and integrate \(e^{3x}\small.\)

$$ \begin{matrix} u'=e^{3x} \:&\: v=x^2 \\ u=\large\frac{1}{3}\normalsize e^{3x} \:&\: v'=2x \\ \end{matrix} $$

For brevity, let \(I=\Large\int\normalsize\!x^2\,e^{3x}\,dx\small.\)

Now we apply integration by parts:

$$ \begin{eqnarray} I &=& uv-\int\,uv'\,dx \\[6pt] &=& \small\frac{1}{3}\normalsize e^{3x}.x^2-\int \small\frac{1}{3}\normalsize e^{3x}.2x\,dx \\[6pt] &=& \small\frac{1}{3}\normalsize x^2\,e^{3x}-\small\frac{2}{3}\normalsize\int e^{3x}.x\,dx \end{eqnarray} $$

Now we need to integrate by parts again, this time using the following:

$$ \begin{matrix} u'=e^{3x} \:&\: v=x \\ u=\large\frac{1}{3}\normalsize e^{3x} \:&\: v'=1 \\ \end{matrix} $$

$$ \begin{eqnarray} I &=& \small\frac{1}{3}\normalsize x^2\,e^{3x}-\small\frac{2}{3}\normalsize\left(\small\frac{1}{3}\normalsize e^{3x}\,x-\!\!\int\!\small\frac{1}{3}\normalsize e^{3x}\,dx\right) \\[6pt] &=& \small\frac{1}{3}\normalsize x^2\,e^{3x}-\small\frac{2}{9}\normalsize\,x\,e^{3x}+\!\!\small\frac{2}{9}\normalsize\int\!e^{3x}\,dx \\[6pt] &=& \small\frac{1}{3}\normalsize x^2\,e^{3x}-\small\frac{2}{9}\normalsize\,x\,e^{3x}+\small\frac{2}{27}\normalsize e^{3x}+c \end{eqnarray} $$

First we examine the factors and ask ourselves which we should differentiate and which we should integrate. In this example, it doesn't matter.

$$ \begin{matrix} u'=e^x \:&\: v=cos\,x \\ u=e^x \:&\: v'=-sin\,x \\ \end{matrix} $$

For brevity, let \(I=\Large\int\normalsize\!e^x\,cos\,x\,dx\small.\)

So now we apply integration by parts:

$$ \begin{eqnarray} I &=& uv-\int\,uv'\,dx \\[6pt] &=& e^x\,cos\,x-\int (e^x)\,(-sin\,x)\,dx \\[6pt] &=& e^x\,cos\,x+\int e^x\,sin\,x\,dx \\[6pt] \end{eqnarray} $$

So we have to integrate by parts again, this time using the following:

$$ \begin{matrix} u'=e^x \:&\: v=sin\,x \\ u=e^x \:&\: v'=cos\,x \\ \end{matrix} $$

What is going to happen here is that \(I\) will appear again on the right, so we will add it to both sides and solve from there.

$$ \begin{eqnarray} I &=& e^x\,cos\,x+\left(e^x\,sin\,x-\!\!\int\!e^x\,cos\,x\,dx\right) \\[6pt] &=& e^x\,cos\,x+e^x\,sin\,x-I \\[6pt] 2I&=& e^x\,cos\,x+e^x\,sin\,x \\[6pt] I&=& \small\frac{1}{2}\normalsize\left(e^x\,cos\,x+e^x\,sin\,x\right)+c \\[6pt] \end{eqnarray} $$

We are being asked to integrate a product, so this example requires integration by parts:

$$ \int\,u'v\,dx = uv-\int\,uv'\,dx $$

We start by examining the two factors, \(x^7\) and \((ln\,x)^{2}\small,\) and asking ourselves which we should differentiate and which we should integrate. Clearly we shouldn't differentiate \(x^7\) because its power would only reduce by one each time we differentiate, and also you haven't yet seen a standard integral for \(ln\,x\small,\) let alone \((ln\,x)^{2}\small.\) So we will differentiate \((ln\,x)^2\small\) and hope that it simplifies.

$$ \begin{matrix} u'=x^7 \:&\: v=(ln\,x)^2 \\ u=\frac{1}{8}x^8 \:&\: v'=2(ln\,x)(\frac{1}{x}) \\ \end{matrix} $$

For brevity, let \(I=\Large\int\normalsize\!x^{7}\,(ln\,x)^{2}\,dx\small.\)

So now we apply integration by parts:

$$ \begin{eqnarray} I &=& \small\frac{1}{8}\normalsize x^8(ln\,x)^2-\int\,\small\frac{1}{8}\normalsize x^{8}.2(ln\,x)\small\left(\frac{1}{x}\right)\normalsize\,dx \\[6pt] &=& \small\frac{1}{8}\normalsize x^8(ln\,x)^2-\small\frac{1}{4}\normalsize \int\,x^{7}(ln\,x)\,dx \\[6pt] &=& \small\frac{1}{8}\normalsize x^8(ln\,x)^2-\small\frac{1}{4}\normalsize J\small\textsf{, say.} \\[6pt] \end{eqnarray} $$

Integrating what we have called \(J\) here requires a second application of integration by parts. So, reusing the \(u,\,v\) notation:

$$ \begin{matrix} u'=x^7 \:&\: v=ln\,x \\ u=\frac{1}{8}x^8 \:&\: v'=\frac{1}{x} \\ \end{matrix} $$

You will notice that \(v'\) is much simpler this time. That's a good sign!

$$ \begin{eqnarray} J &=& \small\frac{1}{8}\normalsize x^8\,ln\,x-\int\small\frac{1}{8}\normalsize x^{8}\small\left(\frac{1}{x}\right)\normalsize\,dx \\[6pt] &=& \small\frac{1}{8}\normalsize x^{8}\,ln\,x-\small\frac{1}{8}\normalsize \int x^7\,dx \\[6pt] &=& \small\frac{1}{8}\normalsize x^{8}\,ln\,x-\small\frac{1}{8}\normalsize \left(\small\frac{1}{8}\normalsize x^8\right)+c \\[6pt] &=& \small\frac{1}{8}\normalsize x^{8}\,ln\,x-\small\frac{1}{64}\normalsize x^8+c \\[6pt] \end{eqnarray} $$

Subsituting \(J\) back into \(I\small,\) we obtain:

$$ \begin{eqnarray} I &=& \small\frac{1}{8}\normalsize x^8(ln\,x)^2\!-\small\frac{1}{4}\normalsize \left(\!\small\frac{1}{8}\normalsize x^{8}\,ln\,x\!-\!\small\frac{1}{64}\normalsize x^{8}\!\right)\!+\!c \\[6pt] &=& \small\frac{1}{8}\normalsize x^8(ln\,x)^2-\small\frac{1}{32}\normalsize x^{8}\,ln\,x+\small\frac{1}{256}\normalsize x^{8}+c \\[6pt] \end{eqnarray} $$

From Higher, the equation of a circle radius \(r\) centred on the origin is \(x^2+y^2=r^2\small.\)

Rearranging, \(y^2=r^2-x^2\small.\)

So \(y=\sqrt{r^2-x^2}\) is the upper semicircle, which we will now rotate \(2\pi\) radians around the \(x\)-axis to form a sphere. The lower limit of \(\raise 0.2pt{x}\) is \(\raise 0.2pt{-r}\) and the upper limit is \(\raise 0.2pt{r\small.}\)

$$ \begin{eqnarray} V &=& \int^{r}_{-r}\pi y^{2}\,dx \\[6pt] &=& \int^{r}_{-r}\pi (r^2-x^2)\,dx \\[6pt] &=& \pi\left[r^{2}x-\small\frac{1}{3}\normalsize x^3\right]^{\phantom{.}r}_{-r} \\[6pt] &=& \pi\left[\left(r^{3}-\small\frac{1}{3}\normalsize r^3\right)-\left(-r^3+\small\frac{1}{3}\normalsize r^3\right)\right] \\[6pt] &=& \pi\left(r^{3}-\small\frac{1}{3}\normalsize r^3+r^3-\small\frac{1}{3}\normalsize r^3\right) \\[6pt] &=& \pi\left(2r^{3}-\small\frac{2}{3}\normalsize r^3\right) \\[6pt] &=& \pi\left(\small\frac{6}{3}\normalsize r^{3}-\small\frac{2}{3}\normalsize r^3\right) \\[6pt] &=& \small\frac{4}{3}\normalsize \pi r^3 \\[6pt] \end{eqnarray} $$

In this question, the curve is to be rotated around the \(y\)-axis, not the \(x\)-axis.

So the required volume is given by:

where \(L\) and \(U\) are the lower and upper limits. So we need to find these limits and an expression for \(x^{2}\small.\)

As the limits are \(y\)-values, we solve for \(x\!=\!0\small.\)

$$ \begin{gather} 4x^2+9y^2=36\\[6pt] 4(0)^2+9y^2=36 \\[6pt] 9y^2=36 \\[6pt] y^2=4 \\[6pt] y=\pm\,2 \end{gather} $$

However, we are only rotating the upper-right quadrant of this curve, so the lower limit is \(0\small,\) not \(-2\small.\)

Now we rearrange to obtain an expression for \(x^{2}\small.\)

$$ \begin{gather} 4x^2+9y^2=36\\[6pt] 4x^2=36-9y^2 \\[6pt] x^2=\small\frac{1}{4}\normalsize(36-9y^2) \end{gather} $$

We are now ready to integrate to find the volume of revolution.

$$ \begin{eqnarray} V &=& \int^{2}_{0}\pi x^{2}\,dy \\[8pt] &=& \small\frac{\pi}{4}\normalsize\int^{2}_{0}(36-9y^2)\,dy \\[8pt] &=& \small\frac{\pi}{4}\normalsize\left[36y-\small\frac{9}{3}\normalsize y^3\right]^{2}_{0} \\[8pt] &=& \small\frac{\pi}{4}\normalsize\left[36y-3y^3\right]^{2}_{0} \\[8pt] &=& \small\frac{\pi}{4}\normalsize\left[\left(36(2)\!-\!3(2^3)\right)-\left(36(0)\!-\!3(0^3)\right)\right] \\[8pt] &=& \small\frac{\pi}{4}\normalsize\left(72-3(8)\right) \\[8pt] &=& \small\frac{\pi}{4}\normalsize\left(72-24\right) \\[8pt] &=& \small\frac{\pi}{4}\normalsize\left(48\right) \\[8pt] &=& 12\pi\ \small\textsf{cubic units} \end{eqnarray} $$

(a)  We start by examining the two factors, \((x^2\!-\!2x\!+\!1)\) and \(e^{4x}\small,\) and asking ourselves which we should differentiate and which we should integrate. Clearly it doesn't matter for \(e^{4x}\) and the quadratic should be differentiated to reduce its degree.

So now we apply integration by parts, twice:

$$ \begin{eqnarray} &\phantom{=}&\int^{\small{1}}_{\small{0}}\underbrace{(x^2-2x+1)}_{\textsf{u}}\underbrace{e^{4x}}_{\textsf{v'}}\,dx\\[8pt] &=& \large\left[\normalsize \left(x^2\!-\!2x\!+\!1\right)\tiny\left(\small\frac{1}{4}\normalsize e^{4x}\tiny\right)\large\right]^{\normalsize1}_{\normalsize0}-\normalsize\int^{\small{1}}_{\small{0}}(2x\!-\!2)\tiny\left(\small\frac14\normalsize e^{4x}\tiny\right)\normalsize\,dx\\[8pt] &=& \left(0-\small\frac14\right)\normalsize-\small\frac12\normalsize\int^{\small{1}}_{\small{0}}\underbrace{(x\!-\!1)}_{\textsf{u}}\underbrace{e^{4x}}_{\textsf{v'}}\,dx\\[8pt] &=& -\!\small\frac14-\small\frac12\large\left[\normalsize(x\!-\!1)(\small\frac{1}{4}\normalsize e^{4x})- \int^{\small{1}}_{\small{0}}1.\small\frac{1}{4}\normalsize e^{4x}\,dx\large\right]^{\normalsize1}_{\normalsize0}\\[8pt] &=& -\!\small\frac14-\small\frac12\large\left[\normalsize(x\!-\!1)(\small\frac{1}{4}\normalsize e^{4x})-\small\frac{1}{16}\normalsize e^{4x}\large\right]^{\normalsize1}_{\normalsize0} \\[8pt] &=& -\!\small\frac14-\small\frac12\normalsize\left(\small\left(\normalsize0-\small\frac{1}{16}\normalsize e^4\small\right)\normalsize-\small\left(\normalsize-\small\frac14\normalsize-\small\frac{1}{16}\normalsize\small\right)\normalsize\right)\\[8pt] &=& -\!\small\frac14-\small\frac12\normalsize\left(-\small\frac{1}{16}\normalsize e^4+\small\frac{5}{16}\normalsize\right)\\[8pt] &=& -\!\small\frac14+\small\frac{1}{32}\normalsize e^4-\small\frac{5}{32}\normalsize\\[8pt] &=& \small\frac{1}{32}\normalsize e^4-\small\frac{13}{32}\normalsize\\[8pt] &=& \small\frac{1}{32}\normalsize\left(e^4-13\right)\\[8pt] \end{eqnarray} $$

(b)  This part makes use of the answer to part (a) to shortcut the integration process.

$$ \begin{eqnarray} V &=& \int^{1}_{0}\pi y^{2}\,dx \\[8pt] &=& \pi\int^{1}_{0}\{4(x\!-\!1)e^{2x}\}^2\,dx \\[8pt] &=& 16\pi\int^{1}_{0} \{(x^2\!-\!2x\!+\!1)e^{4x}\}^2\,dx \\[8pt] &=& \small\frac{16\pi}{32}\normalsize\left(e^4-13\right)\\[8pt] &=& \small\frac{\pi}{2}\normalsize\left(e^4-13\right)\ \small\textsf{cubic units} \end{eqnarray} $$