2nd method to evaluate the indefinite integral using must know basic techniques

[@user-lu6yg3vk9z]

Another method
sin^2(x)/(sin^4(x)-1)
sin^4(x)/(sin^2(x)(sin^4(x)-1))
(sin^4(x)-1+1)/(sin^2(x)(sin^4(x)-1))
(sin^4(x)-1)/(sin^2(x)(sin^4(x)-1)) +1/(sin^2(x)(sin^4(x)-1))
1/sin^2(x)+1/(sin^2(x)(sin^2(x)+1)(sin^2-1))
csc^2(x)-1/(sin^2(x)(sin^2(x)+1)(1-sin^2(x))
csc^2(x)-1/(sin^2(x)(sin^2(x)+1)cos^2(x))
csc^2(x)-sec^2(x)/(sin^2(x)(sin^2(x)+1))
csc^2(x)-sec^4(x)sec^2(x)/(tan^2(x)(tan^2(x)+sec^2(x))
csc^2(x)-(tan^2(x)+1)^2sec^2(x)/(tan^2(x)(2tan^2(x)+1))
Let u=tan(x)
du=sec^2(x) dx
csc^2(x)-(u^2+1)^2/(u^2(2u^2+1))
csc^2(x)-(u^4+2u^2+1)/(u^2(2u^2+1))
csc^2(x)-u^4/(u^2(2u^2+1)+(2u^2+1)/(u^2(2u^2+1))
csc^2(x)-u^2/(2u^2+1)+1/u^2
csc^2(x)-1/2 integral 2u^2+1-1/(2u^2+1)+1/u^2
csc^2(x)-1/2integral 1-1/(2u^2+1)+1/u^2

[@cipherunity]

It is done. You may watch the video for your second solution for this integral