Question #581c1

1 Answer
Bill K.
Jun 10, 2015

#int (sin^6(x) + cos^6(x))/ (sin^2(x) cos ^2(x))\ dx=tan(x)-cot(x)-3x+C#

Explanation:

First note that #sin^6(x)+cos^6(x)# is a sum of two cubes #(sin^2(x))^3+(cos^2(x))^3# so it can be factored using #a^3+b^3=(a+b)(a^2-ab+b^2)# and the Pythagorean identity to get

#sin^6(x)+cos^6(x)=(sin^2(x))^3+(cos^2(x))^3#

#=(sin^2(x)+cos^2(x))(sin^4(x)-sin^2(x)cos^2(x)+cos^4(x))#

#=sin^4(x)-sin^2(x)cos^2(x)+cos^4(x)#.

Therefore,

# (sin^6 x + cos^6 x)/ (sin^2(x) cos ^2(x))= (sin^4(x)-sin^2(x)cos^2(x)+cos^4(x))/ (sin^2(x) cos ^2(x))#

#=tan^2(x)-1+cot^2(x)#

But #tan^2(x)=sec^2(x)-1# and #cot^2(x)=csc^2(x)-1#. Hence,

# (sin^6 x + cos^6 x)/ (sin^2(x) cos ^2(x))=sec^2(x)+csc^2(x)-3# and thus

#int (sin^6(x) + cos^6(x))/ (sin^2(x) cos ^2(x))\ dx=int sec^2(x)+csc^2(x)-3\ dx#

#=tan(x)-cot(x)-3x+C#

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