STEP I 2004 4

Using the quotient rule:

\begin{align*}(\sec t)' & = \left(\frac 1 {\cos t}\right)' \\ & = \frac {-(-\sin t)} {\cos^2 t} \\ & = \frac {\sin t} {\cos t} \times \frac 1 {\cos t} \\ & = \tan t \sec t\end{align*}

(i)

We have if x = \sqrt 2, then \cos t = \dfrac 1 {\sqrt 2}, so t = \dfrac \pi 4. Similarly for x = 2 we have \cos t = \dfrac 1 2 so t = \dfrac \pi 3, so we have:

\begin{align*}\int_{\sqrt 2}^2 \frac 1 {x^3 \sqrt {x^2 - 1}} \mathrm dx & = \int_{\pi/4}^{\pi/3} \frac {\sec t \tan t} {(\sec t)^3 \sqrt {\sec^2 t - 1}} \mathrm dt \\ & = \int_{\pi/4}^{\pi/3} \frac {\sec t \tan t} {\sec^3 t \sqrt {\tan^2 t}} \mathrm dt \\ & = \int_{\pi/4}^{\pi/3} \cos^2 t \mathrm dt\end{align*}

having used 1 + \tan^2 t \equiv \sec^2 t from the first to second line.

We then have:

\begin{align*}\int_{\pi/4}^{\pi/3} \cos^2 t \mathrm dt & = \frac 1 2 \int_{\pi/4}^{\pi/3} (1 + \cos 2t)\mathrm dt \\ & = \frac 1 2 \left[t + \frac 1 2 \sin 2t\right]_{\pi/4}^{\pi/3} \\ & = \frac 1 2 \left(\left(\frac \pi 3 - \frac \pi 4\right) + \frac 1 2 \sin \frac {2 \pi} 3 - \frac 1 2 \sin \frac \pi 2\right) \\ & = \frac 1 2 \left(\frac \pi {12} + \frac {\sqrt 3} {4} - \frac 1 2\right) \\ & = \frac 1 2 \left(\frac \pi {12} + \frac {\sqrt 3 - 2} 4\right) \\ & = \frac \pi {24} + \frac {\sqrt 3 - 2} 8\end{align*}

We want to try reducing the next two integrals into a form similar to this one, namely:

\displaystyle \int \frac 1 {x^n \sqrt {x^2 - a^2}} \mathrm dx

which we know reduces nicely with x = a \sec t.

(ii)

Set x + 2 = u, then x + 1 = u - 1 and x + 3 = u + 1, so:

\begin{align*} \int \frac 1 {(x + 2) \sqrt {(x + 1)(x + 3)}} \mathrm dx &= \int \frac 1 {u \sqrt {(u - 1)(u + 1)}} \mathrm du \\ & = \int \frac 1 {u \sqrt {u^2 - 1}} \mathrm du\end{align*}

[alternatively notice that (x + 1)(x + 3) = (x + 2)^2 - 1, from which the substitution leads on more naturally]

Setting u = \sec t as before we have:

\begin{align*}\int \frac 1 {u \sqrt {u^2 - 1}} \mathrm du & = \int \frac {\sec t \tan t} {\sec t \tan t} \mathrm dt \\ & = t + C \\ & = \operatorname{arcsec}(u) + C \\ & = \operatorname{arcsec}(x + 2) + C \end{align*}

You can also write this:

\displaystyle \arccos \left(\frac 1 {x + 2}\right) + C

(iii)

Write:

\displaystyle \frac 1 {(x + 2) \sqrt {x^2 + 4x - 5}} = \frac 1 {(x + 2) \sqrt {(x + 2)^2 - 9}}

(completing the square)

Setting 3u = x + 2 we have:

\begin{align*}\int \frac 1 {(x + 2) \sqrt {x^2 + 4x - 5}} \mathrm dx &= \int \frac 3 {3u \sqrt {(3u)^2 - 9}} \mathrm du \\ & = \frac 3 9 \int \frac 1 {u \sqrt {u^2 - 1}} \mathrm du \\ & = \frac 1 3 \operatorname{arcsec}(u) + C \\ & = \frac 1 3 \operatorname{arcsec}\left(\frac {x + 2} 3\right) + C\end{align*}

using the result from part (ii)

Similarly you could write this:

\displaystyle \frac 1 3 \arccos \left(\frac 3 {x + 2}\right) + C

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