Hatena::ブログ(Diary)

数検に挑戦!

2008-09-16

数検1級一次過去問(5)

00:00

【問題】

Find the following indefinite integral.

  ¥int sin^4xdx

【解答】

  ¥int sin^4xdx = ¥int sin^2x(1-cos^2x)dx

          =¥int sin^2xdx - ¥int sin^2xcos^2xdx

          =¥frac12¥int(1-cos2x)dx - ¥frac14¥int(1-cos2x)(1+cos2x)dx

          =¥frac12(x-¥frac12sin2x) - ¥frac14¥int(1-cos^22x)dx

          =¥frac12(x-¥frac12sin2x) - ¥frac14¥int(1-¥frac{1+cos4x}2)dx

          =¥frac12(x-¥frac12sin2x) - ¥frac14x + ¥frac18¥int(1+cos4x)dx

          =¥frac12(x-¥frac12sin2x) - ¥frac14x + ¥frac18(x+¥frac14sin4x) + C

          =¥frac38x-¥frac14sin2x + ¥frac1{32}sin4x + C