What is the integral of 1/(1-sin^4(x)) ?

The integral of 1/(1-sin^4(x)) is 1/(2sqrt(2))arctan(sqrt(2)tan(x))+1/2 tan(x)+C

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integral of 1/(1-sin^4(x))

\int \frac{1}{1 - sin ^{4} ( x )} d x

\frac{1}{2 2} arctan (2 tan (x)) + \frac{1}{2} tan (x) + C

Solution steps

\int \frac{1}{1 - sin ^{4} ( x )} d x

Apply Trigonometric Substitution

: \int \frac{1 + u ^{2}}{1 + 2 u ^{2}} d u

= \int \frac{1 + u ^{2}}{1 + 2 u ^{2}} d u

Expand

\frac{1 + u ^{2}}{1 + 2 u ^{2}} : \frac{1}{1 + 2 u ^{2}} + \frac{u ^{2}}{1 + 2 u ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{1 + 2 u ^{2}} d u + \int \frac{u ^{2}}{1 + 2 u ^{2}} d u

\int \frac{1}{1 + 2 u ^{2}} d u = \frac{1}{2} arctan (2 u)

\int \frac{u ^{2}}{1 + 2 u ^{2}} d u = - \frac{1}{2 2} arctan (2 u) + \frac{1}{2} u

= \frac{1}{2} arctan (2 u) - \frac{1}{2 2} arctan (2 u) + \frac{1}{2} u

Substitute back

u = tan (x)

= \frac{1}{2} arctan (2 tan (x)) - \frac{1}{2 2} arctan (2 tan (x)) + \frac{1}{2} tan (x)

Add similar elements:

\frac{1}{2} arctan (2 tan (x)) - \frac{1}{2 2} arctan (2 tan (x)) = \frac{1}{2 2} arctan (2 tan (x))

= \frac{1}{2 2} arctan (2 tan (x)) + \frac{1}{2} tan (x)

Add a constant to the solution

= \frac{1}{2 2} arctan (2 tan (x)) + \frac{1}{2} tan (x) + C

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What is the integral of 1/(1-sin^4(x)) ?
The integral of 1/(1-sin^4(x)) is 1/(2sqrt(2))arctan(sqrt(2)tan(x))+1/2 tan(x)+C