STEP III 1994 question 1 solution

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STEP III 1994 question 1 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 1994 question 1 solution

$\\ \displaystyle \int_{0}^{x} sech (t) \text{d}t\\=\int_{0}^{x}\frac{\cosh t}{\cosh^2 t} \text{d}t\\=\int_{0}^{x}\frac{\cosh t}{1+\sinh^2 t}\texb{d}t$

Using the substitution $u=\sinh t$

$\displaystyle \\=\int_{0}^{\sinh x}\frac{1}{1+u^2}\text{d}u\\=\arctan(\sinh x)$

To find the reduction formula:

$\displaystyle \\I_{n}=\int_{0}^{x} sech^n(t)\text{d}t\\=\int_{0}^{x}(sech^2(t))(sech^{n-2}(t))\text{d}t\\=\int_{0}^{x}(1-\tanh^2 t)(sech^{n-2}(t))\text{d}t\\=\int_{0}^{x}sech^{n-2}(t)\text{d}t-\int_{0}^{x}(\tanh t)(\tanh t \times sech^{n-2}(t))\text{d}t$

Using integration by parts,

$\displaystyle \\I_{n}=I_{n-2}+[\frac{\tanh(t)\times sech^{n-2}(t)}{n-2}]_{0}^{x}-\int_{0}^{x}\frac{sech^{2}(t)\times sech^{n-2}(t)}{n-2}\text{d}t\\ \therefore I_{n}=(\frac{n-2}{n-1})I_{n-2}+(\frac{1}{n-1})\tanh(x)\times sech^{n-2}(x)$

So,

$\displaystyle \\I_{5}=\frac{3}{8}I_{1}+\frac{3\tanh(x)sech(x)}{8}+\frac{\tanh(x)sech^{3}(x)}{4}\\ I_{5}=\frac{3}{8}\arctan(\sinh x) +\frac{3\tanh(x)sech(x)}{8}+\frac{\tanh(x)sech^{3}(x)}{4}$

and,

$\displaystye \\I_{6}=\frac{8}{15}I_{2}+\frac{4}{15}\tanh(x)sech^{2}(x)+\frac{1}{5}\tanh(x)sech^{4}(x)\\ I_{6}=\frac{8}{15}\tanh x+\frac{4}{15}\tanh(x)sech^{2}(x)+\frac{1}{5}\tanh(x)sech^{4}(x)$