 | # (1)は導関数の計算のみ、(2)は積分計算のみすれば良いのですよね?
(1)
y = {sin(x)}/{((a*cos(x))^2+(b*sin(x))^2)^(1/2)}と解釈して回答します。
y = f(x)/g(x)のとき、dy/dx = (f'(x)*g(x)-f(x)*g'(x))/(g(x)^2)です。
f(x) = sin(x), g(x) = ((a*cos(x))^2+(b*sin(x))^2)^(1/2)
f'(x) = cos(x)
g'(x) = (1/2)*{((a*cos(x))^2+(b*sin(x))^2)^(-1/2)}*{2*a^2*cos(x)*(-sin(x))+2*b^2*sin(x)*cos(x)}
= sin(x)*cos(x)*{b^2-a^2}/g(x)
f'(x)*g(x)-f(x)*g'(x) = cos(x)*g(x)-(sin(x)^2)*cos(x)*(b^2-a^2)/g(x)
= {cos(x)*g(x)^2-(sin(x)^2)*cos(x)*(b^2-a^2)}/g(x)
= {cos(x)*((a*cos(x))^2+(b*sin(x))^2)-(sin(x)^2)*cos(x)*(b^2-a^2)}/g(x)
= {a^2*cos(x)^3+b^2*cos(x)*sin(x)^2-b^2*sin(x)^2*cos(x)+a^2*sin(x)^2*cos(x)}/g(x)
= {a^2*cos(x)*(cos(x)^2+sin(x)^2)}/g(x)
= {a^2*cos(x)}/g(x)
dy/dx = (f'(x)*g(x)-f(x)*g'(x))/(g(x)^2) = a^2*cos(x)/(g(x)^3)
= a^2*cos(x)/{((a*cos(x))^2+(b*sin(x))^2)^(3/2)}
(2)
I = ∫{(log[e](x))^2}dxとおきます。
部分積分により、
I = x*(log[e](x))^2-∫{x*2*log[e](x)*(1/x)}dx
= x*(log[e](x))^2-2*∫{log[e](x)}dx
= x*(log[e](x))^2-2*{x*log[e](x)-∫{x*(1/x)}dx}
= x*(log[e](x))^2-2x*log[e](x)+2*∫{x*(1/x)}dx
= x*(log[e](x))^2-2x*log[e](x)+2*∫dx
= x*(log[e](x))^2-2x*log[e](x)+2x
# 積分定数は省略しています。
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