Question

Int1/1-sin^4x =?

Answer 1

`I=1/4(2tan(x)+sqrt(2)arctan(sqrt(2)tan(x))+C`

Explanation:

We want to solve

`I=int1/(1-sin^4(x))dx`

Multiply the numerator and denominator by `sec^4(x)`

`I=intsec^4(x)/(sec^4(x)-tan^4(x))dx`

This may seems like a random idea,
but often if we can make the integrand involve tangens and secant,
this leads to good substitution opportunities, as we will see

Make a substitution `u=tan(x)=>du=sec^2(x)dx`

`I=intsec^2(x)/(sec^4(x)-u^4)du`

Remember `color(blue)(sec^2(x)=tan^2(x)+1`

`I=int(u^2+1)/((u^2+1)^2-u^4)du`

`color(white)(I)=int(u^2+1)/(2u^2+1)du`

By long division

`I=1/2int1du+1/2int1/(2u^2+1)du`

Make a substitution `s=sqrt(2)u=>ds=sqrt(2)du`

`I=1/2int1du+1/2^(3/2)int1/(s^2+1)ds`

`color(white)(I)=1/2u+1/2^(3/2)arctan(s)+C`

`color(white)(I)=1/4(2u+sqrt(2)arctan(s))+C`

Substitute back `s=sqrt(2)u` and `u=tan(x)`

`I=1/4(2tan(x)+sqrt(2)arctan(sqrt(2)tan(x))+C`