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STEP III 1994 question 1 solutionTSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 1994 question 1 solution
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\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
Solution by khaixiang |
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![\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)](http://www.thestudentroom.co.uk/latexrender/pictures/ff/ff9e09bd0da5d90a997bad8f7f3953f3.png "\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)")
