How do you find the integral of #sin^2(x)cos^4(x) #?

Question

Ezra Adams · Accepted Answer

#int sin^2(x)cos^4(x) dx#

#= -1/6 sin^5(x) cos(x)+7/24 sin^3(x) cos(x)-1/16 sin(x) cos(x)+1/16 x + C#

First note that: #sin^2(x)cos^4(x) = (1 - cos^2(x)) cos^4(x) = -cos^(6)x + cos^4(x)# #d/(dx) sin^5(x) cos(x) = 5 sin^4(x) cos^2(x) - sin^6(x)# #color(white)(d/(dx) sin^5(x) cos(x)) = sin^4(x)(5 cos^2(x) - sin^2(x))# #color(white)(d/(dx) sin^5(x) cos(x)) = (1-cos^2(x))(1-cos^2(x))(5 cos^2(x) - (1-cos^2(x)))# #color(white)(d/(dx) sin^5(x) cos(x)) = (cos^4(x)-2cos^2(x)+1)(6 cos^2(x) - 1)# #color(white)(d/(dx) sin^5(x) cos(x)) = 6cos^6(x)-13cos^4(x)+8cos^2(x)-1# #d/(dx) sin^3(x) cos(x) = 3 sin^2(x) cos^(2)(x) - sin^4(x)# #color(white)(d/(dx) sin^3(x) cos(x)) = sin^2(x)(3 cos^2(x) - sin^2(x))# #color(white)(d/(dx) sin^3(x) cos(x)) = (1-cos^2(x))(3 cos^2(x) - (1-cos^2(x)))# #color(white)(d/(dx) sin^3(x) cos(x)) = (-cos^2(x)+1)(4 cos^2(x) - 1)# #color(white)(d/(dx) sin^3(x) cos(x)) = -4cos^4(x)+5cos^2(x)-1# #d/(dx) sin(x) cos(x) = cos^2(x)-sin^2(x) = 2cos^2(x) - 1# Now we can choose multipliers to make the running sum match each coefficient of #cos^(2k)(x)# in descending order: #d/(dx) (-1/6 sin^5(x) cos(x)) = -cos^6(x)+13/6 cos^4(x)-4/3 cos^2(x)+1/6# #d/(dx) (7/24 sin^3(x) cos(x)) = -7/6 cos^4(x)+35/24 cos^2(x) - 7/24# #d/(dx) (-1/16 sin(x) cos(x)) = -1/8 cos^2(x) + 1/16# #d/(dx) (1/16 x) = 1/16# summing to: #-cos^6(x)+cos^4(x) = sin^2(x)cos^4(x)# So: #int sin^2(x)cos^4(x) dx# #= -1/6 sin^5(x) cos(x)+7/24 sin^3(x) cos(x)-1/16 sin(x) cos(x)+1/16 x + C#

Elijah Abram · Answer

undefined To find the integral of sin^2(x)cos^4(x), you can use trigonometric identities to simplify the expression. Start by using the identity sin^2(x) = 1 - cos^2(x), then apply the power-reducing formula cos^2(x) = (1 + cos(2x))/2. After simplifying, you'll have an expression that can be integrated more easily using basic integration techniques.