# Int1/1-sin^4x =?

##### 1 Answer

#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

#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

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

Remember

#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

#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

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