Integral of the sine function with exponent solve the following integral: int sin ^ 6xdx?

2 Answers
Mar 29, 2018

#-1/6sin^5xcosx-5/24sin^3xcos x -5/16 sin x cos x+5/16 x+C#

Explanation:

Instead of solving the specific integral, it will be simpler to derive a recursion formula for the general integral

#color(red)(I_(2n) equiv int sin^(2n) x dx)#

we will also suppress the constant of integration till the end.

Try integration by parts with #sin^(2n-1) x# as the first function, and #sin x# as the second.

#I_{2n} equiv int sin^(2n) x dx = sin^(2n-1)x int sin x dx#
#qquad -int [(d/dx sin^(2n-1) x)times int sin x dx]dx#
#qquad = -sin^(2n-1) x cos x#
#qquad -int [((2n-1)sin^(2n-2)xcos x)times (-cos x)] dx#
#qquad = -sin^(2n-1) x cos x+int (2n-1)sin^(2n-2)xcos^2 x dx#
#qquad = -sin^(2n-1) x cos x#
#qquad +int (2n-1)sin^(2n-2)x(1-sin^2 x) dx#
#qquad = -sin^(2n-1) x cos x+ (2n-1)(I_(2n-2)-I_(2n))#

This can be rearranged to write

#2n I_(2n) = -sin^(2n-1) x cos x+ (2n-1)I_(2n-2)#

so that

#I_(2n) = -1/(2n) sin^(2n-1) x cos x+ (2n-1)/(2n) I_(2n-2)#

This gives a recursion relation that can be used to find #I_(2n)#. Note that #I_0 = int sin^0 x dx = x#

In our case, we are looking for #I_6#

#I_6 = -1/6sin^5xcosx+5/6I_4#
#qquad =-1/6sin^5xcosx+5/6(-1/4sin^3xcos x+3/4I_2)#
#qquad =-1/6sin^5xcosx-5/24sin^3xcos x #
#qquad quad +5/8(-1/2sin x cos x+1/2x)#
#qquad = -1/6sin^5xcosx-5/24sin^3xcos x #
#qquad qquad -5/16 sin x cos x+5/16 x#

Mar 29, 2018

An alternative version of the answer is

# -1/192 sin(6x) +3/64 sin(4x)-15/64cos(2x)+5/16 x+C#

Explanation:

Using #sinx =( e^{ix}-e^{-ix})/(2i)# we get

#sin^6x =(( e^{ix}-e^{-ix})/(2i))^6 #
#qquad = 1/(2i)^6[(e^(ix))^6 + 6(e^(ix))^5(-e^(-ix))+15 (e^(ix))^4(-e^(-ix))^2+20(e^(ix))^3(-e^(-ix))^3+15(e^(ix))^2(-e^(-ix))^4+6(e^(ix))^1(-e^(-ix))^5+(-e^(-ix))^6]#
# qquad = -1/64[(e^(i6x)+e^(-i6x))-6(e^(i4x)+e^(-i4x)) + 15(e^(i2x)+e^-(i2x))-20]#
# = -1/32 cos(6x) +3/16 cos(4x) -15/32 cos(2x)+5/16#

This is easily integrated to yield

# int sin^6 x dx = int (-1/32 cos(6x) +3/16 cos(4x) -15/32 cos(2x)+5/16) dx#
#qquad = -1/192 sin(6x) +3/64 sin(4x)-15/64cos(2x)+5/16 x+C#