Int1/1-sin^4x =?

Apr 1, 2018

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

Explanation:

We want to solve

$I = \int \frac{1}{1 - {\sin}^{4} \left(x\right)} \mathrm{dx}$

Multiply the numerator and denominator by ${\sec}^{4} \left(x\right)$

$I = \int {\sec}^{4} \frac{x}{{\sec}^{4} \left(x\right) - {\tan}^{4} \left(x\right)} \mathrm{dx}$

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 $u = \tan \left(x\right) \implies \mathrm{du} = {\sec}^{2} \left(x\right) \mathrm{dx}$

$I = \int {\sec}^{2} \frac{x}{{\sec}^{4} \left(x\right) - {u}^{4}} \mathrm{du}$

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

$I = \int \frac{{u}^{2} + 1}{{\left({u}^{2} + 1\right)}^{2} - {u}^{4}} \mathrm{du}$

$\textcolor{w h i t e}{I} = \int \frac{{u}^{2} + 1}{2 {u}^{2} + 1} \mathrm{du}$

By long division

$I = \frac{1}{2} \int 1 \mathrm{du} + \frac{1}{2} \int \frac{1}{2 {u}^{2} + 1} \mathrm{du}$

Make a substitution $s = \sqrt{2} u \implies \mathrm{ds} = \sqrt{2} \mathrm{du}$

$I = \frac{1}{2} \int 1 \mathrm{du} + \frac{1}{2} ^ \left(\frac{3}{2}\right) \int \frac{1}{{s}^{2} + 1} \mathrm{ds}$

$\textcolor{w h i t e}{I} = \frac{1}{2} u + \frac{1}{2} ^ \left(\frac{3}{2}\right) \arctan \left(s\right) + C$

$\textcolor{w h i t e}{I} = \frac{1}{4} \left(2 u + \sqrt{2} \arctan \left(s\right)\right) + C$

Substitute back $s = \sqrt{2} u$ and $u = \tan \left(x\right)$

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