解答
∫cos7(xta)n3xdx
解答
3675a2t2n3(3675atxsin(atx)−3675atxsin3(atx)+2205atxsin5(atx)−525atxsin7(atx)+280cos3(atx)+126cos5(atx)+75cos7(atx)+1680cos(atx))+C
求解步骤
∫cos7(xta)n3xdx
提出常数: ∫a⋅f(x)dx=a⋅∫f(x)dx=n3⋅∫cos7(atx)xdx
使用换元积分法
=n3⋅∫a2t2ucos7(u)du
提出常数: ∫a⋅f(x)dx=a⋅∫f(x)dx=n3a2t21⋅∫ucos7(u)du
使用分布积分法
=n3a2t21(u(sin(u)−sin3(u)+53sin5(u)−71sin7(u))−∫sin(u)−sin3(u)+53sin5(u)−71sin7(u)du)
∫sin(u)−sin3(u)+53sin5(u)−71sin7(u)du=−3516cos(u)+359cos3(u)−253cos5(u)−31cos3(u)+353cos5(u)−491cos7(u)
=n3a2t21(u(sin(u)−sin3(u)+53sin5(u)−71sin7(u))−(−3516cos(u)+359cos3(u)−253cos5(u)−31cos3(u)+353cos5(u)−491cos7(u)))
u=atx代回=n3a2t21(atx(sin(atx)−sin3(atx)+53sin5(atx)−71sin7(atx))−(−3516cos(atx)+359cos3(atx)−253cos5(atx)−31cos3(atx)+353cos5(atx)−491cos7(atx)))
化简 n3a2t21(atx(sin(atx)−sin3(atx)+53sin5(atx)−71sin7(atx))−(−3516cos(atx)+359cos3(atx)−253cos5(atx)−31cos3(atx)+353cos5(atx)−491cos7(atx))):3675a2t2n3(3675atxsin(atx)−3675atxsin3(atx)+2205atxsin5(atx)−525atxsin7(atx)+280cos3(atx)+126cos5(atx)+75cos7(atx)+1680cos(atx))
=3675a2t2n3(3675atxsin(atx)−3675atxsin3(atx)+2205atxsin5(atx)−525atxsin7(atx)+280cos3(atx)+126cos5(atx)+75cos7(atx)+1680cos(atx))
解答补常数=3675a2t2n3(3675atxsin(atx)−3675atxsin3(atx)+2205atxsin5(atx)−525atxsin7(atx)+280cos3(atx)+126cos5(atx)+75cos7(atx)+1680cos(atx))+C