解答
∫sin4(5x)cos6(5x)dx
解答
128001(150x+16cos5(5x)sin(5x)+20cos3(5x)sin(5x)+15sin(10x)−352cos7(5x)sin(5x)+256cos9(5x)sin(5x))+C
求解步骤
∫sin4(5x)cos6(5x)dx
使用换元积分法
=∫sin4(u)cos6(u)51du
提出常数: ∫a⋅f(x)dx=a⋅∫f(x)dx=51⋅∫sin4(u)cos6(u)du
使用三角恒等式改写
=51⋅∫(1−cos2(u))2cos6(u)du
乘开 (1−cos2(u))2cos6(u):cos6(u)−2cos8(u)+cos10(u)
=51⋅∫cos6(u)−2cos8(u)+cos10(u)du
使用积分加法定则: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx=51(∫cos6(u)du−∫2cos8(u)du+∫cos10(u)du)
∫cos6(u)du=6sin(u)cos5(u)+65(41cos3(u)sin(u)+83(u+21sin(2u)))
∫2cos8(u)du=2(8sin(u)cos7(u)+87(6sin(u)cos5(u)+65(41cos3(u)sin(u)+83(u+21sin(2u)))))
∫cos10(u)du=10sin(u)cos9(u)+109(8sin(u)cos7(u)+87(6sin(u)cos5(u)+65(41cos3(u)sin(u)+83(u+21sin(2u)))))
=51(6sin(u)cos5(u)+65(41cos3(u)sin(u)+83(u+21sin(2u)))−2(8sin(u)cos7(u)+87(6sin(u)cos5(u)+65(41cos3(u)sin(u)+83(u+21sin(2u)))))+10sin(u)cos9(u)+109(8sin(u)cos7(u)+87(6sin(u)cos5(u)+65(41cos3(u)sin(u)+83(u+21sin(2u))))))
u=5x代回=51(6sin(5x)cos5(5x)+65(41cos3(5x)sin(5x)+83(5x+21sin(2⋅5x)))−2(8sin(5x)cos7(5x)+87(6sin(5x)cos5(5x)+65(41cos3(5x)sin(5x)+83(5x+21sin(2⋅5x)))))+10sin(5x)cos9(5x)+109(8sin(5x)cos7(5x)+87(6sin(5x)cos5(5x)+65(41cos3(5x)sin(5x)+83(5x+21sin(2⋅5x))))))
化简 51(6sin(5x)cos5(5x)+65(41cos3(5x)sin(5x)+83(5x+21sin(2⋅5x)))−2(8sin(5x)cos7(5x)+87(6sin(5x)cos5(5x)+65(41cos3(5x)sin(5x)+83(5x+21sin(2⋅5x)))))+10sin(5x)cos9(5x)+109(8sin(5x)cos7(5x)+87(6sin(5x)cos5(5x)+65(41cos3(5x)sin(5x)+83(5x+21sin(2⋅5x)))))):128001(150x+16cos5(5x)sin(5x)+20cos3(5x)sin(5x)+15sin(10x)−352cos7(5x)sin(5x)+256cos9(5x)sin(5x))
=128001(150x+16cos5(5x)sin(5x)+20cos3(5x)sin(5x)+15sin(10x)−352cos7(5x)sin(5x)+256cos9(5x)sin(5x))
解答补常数=128001(150x+16cos5(5x)sin(5x)+20cos3(5x)sin(5x)+15sin(10x)−352cos7(5x)sin(5x)+256cos9(5x)sin(5x))+C