STEP II 1992 question 8 solution

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STEP II 1992 question 8 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 1992 question 8 solution

i) $\displaystyle \int \frac{x}{(x-1)(x^2-1)} \ \text{ d} x$

But x^2-1 = (x-1)(x+1)

$\displaystyle \therefore \int \frac{x}{(x-1)^2(x+1)} \ \text{ d} x$

We will now sort this out into partial fractions:

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

Setting x = 1 gives us

$B = \frac{1}{2}$

Setting x = 2 gives us 6A + 2C = 1 (*)

Setting x = 3 gives us 8A + 4C = 1 (#)

Solving (*) and (#) by the elimination method gives us:

$A = \frac{1}{4} \\ C = -\frac{1}{4}$

Therefore we now have to solve

$\displaystyle \int \frac{1}{4(x-1)} + \frac{1}{2(x-1)^2} - \frac{1}{4(x+1)} \ \text{ d} x$

Taking out a constant gives

$\displaystyle \frac{1}{4} \int \frac{1}{(x-1)} + \frac{2}{(x-1)^2} - \frac{1}{(x+1)} \ \text{ d} x = \\ \frac{1}{4} \int \frac{1}{(x-1)} + 2(x-1)^{-2} - \frac{1}{(x+1)} \ \text{ d} x$

Which integrates to

$\frac{1}{4}(\ln(x-1) - \ln(x+1) - \frac{2}{x-1}) + c$

ii. $\displaystyle \int \frac{1}{3\cos x+4\sin x} \text{ d} x$

Let $3\cos x+4\sin x = R\sin(x+ \alpha) \\ = R\sin x\cos\alpha + R\cos x\sin \alpha \ \Rightarrow R = 5 , \ \alpha = \arctan \left(\frac{3}{4}\right)$

So now we are integrating:

$\displaystyle \int \frac{1}{5\sin(x+\arctan\left(\frac{3}{4}\right) )} \ \text{ d} x = \frac{1}{5} \int \csc(x+\arctan\left(\frac{3}{4}\right) ) \ \text{ d} x \\ = -\frac{1}{5}\ln \left|\csc(x+\arctan\left(\frac{3}{4}\right)) + \cot(x+\arctan\left(\frac{3}{4}\right)) \right| + c$

iii. $\displaystyle \int \frac{1}{\sinh x} \ \text{ d} x = \int \frac{2}{e^x-e^{-x}}$

Let $u = e^x \Leftrightarrow \frac{\text{ d} x}{\text{ d}u} = \frac{1}{u}$

So:

$\displaystyle \int \frac{2}{u-\frac{1}{u}} \cdot \frac{1}{u} \ \text{ d}u = \int\frac{2}{u^2-1} \ \text{ d}u$

Using the cover rule for partial fractions gives

$\displaystyle \int \frac{1}{u-1} - \frac{1}{u+1} \text{ d} x = \ln(u-1) - \ln(u+1) + c = \ln\left|\left(\frac{e^x-1}{e^x+1}\right)\right| + c$

Solution by Lusus Naturae.

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