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Expand log(1 + sin2 x) in powers of x as far as the term in x6 |
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Answer» Expand f (x) = log (1 + sin2 x) in powers of x upto x 6 . log (1 + sin2x) = sin2x - \(\frac{sin^4x}{2}\) + \(\frac{sin^6x}{3}\) - ... (∵ log ( 1 + x) = x - \(\frac{x^2}{2}\)+ \(\frac{x^3}{3}\) ... if 1 ≤ x ≤ 1 and here 0 ≤ sin2x ≤ 1) = (x - \(\frac{x^3}{3!}\)+ \(\frac{x^5}{5!}\)- ...)2 - \(\frac{1}{2}\)(x - \(\frac{x^3}{3!}\)+ \(\frac{x^5}{5!}\)- ...)4 + \(\frac{1}{3}\)(x - \(\frac{x^3}{3!}\)+ \(\frac{x^5}{5!}\)- ...)6 - ..... (∵ sinx = x - \(\frac{x^3}{3!}\)+ \(\frac{x^5}{5!}\)- ...) Since, We consider power of x at the terms x6. Therefore, Remove other terms which gives power more than 6. = (x - \(\frac{x^3}{3!}\)+ \(\frac{x^5}{5!}\)+ ...)2 - \(\frac{1}{2}\)(x - \(\frac{x^3}{3!}\)+ \(\frac{x^5}{5!}\)+ ...)4 + (\(\frac{x^6}{3}\) + ....) = (x2 + \(\frac{x^6}{(3!)^2}\) - \(\frac{2x^4}{3!}\) + \(\frac{2x^6}{5!}\) + ....) - \(\frac{1}{2}\)(x4 - \(\frac{4x^6}{3!}\)+...) + (\(\frac{x^6}{3}\) + ...) = x2 - x4(\(\frac{2}{3!}\) + \(\frac{1}{2}\)) + x6(\(\frac{1}{(3!)^2}\)+\(\frac{2}{5!}\)+\(\frac{2}{3!}\)+\(\frac{1}{3}\)) + .... = x2 - x4(\(\frac{2}{6}\) + \(\frac{1}{2}\)) + x6(\(\frac{1}{36}\)+\(\frac{2}{120}\)+\(\frac{2}{6}\)+\(\frac{1}{3}\)) + .... (∵ 3! = 6, 5! = 120) = x2 - x4\((\frac{2+3}{6})\) + x6 \(\times\)\(\frac{10+6+120+120}{360}\) + .... = x2 - \(\frac{5}{6}\)x4 + \(\frac{256}{360}\)x6 +.... = x2 - \(\frac{5}{6}\)x4 + \(\frac{32}{45}\)x6 +.... Hence, Expansion of log(1 + sin2x) in power of x at terms of x6 in log(1+sin2x) = x2 - \(\frac{5}{6}\)x4 + \(\frac{32}{45}\)x6 + .... |
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