What is the integral of (t^3)/(sqrt(1-t^8)) ?

The integral of (t^3)/(sqrt(1-t^8)) is 1/4 arcsin(t^4)+C

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integral of (t^3)/(sqrt(1-t^8))

\int \frac{t ^{3}}{1 - t ^{8}} d t

\frac{1}{4} arcsin (t^{4}) + C

Solution steps

\int \frac{t ^{3}}{1 - t ^{8}} d t

Apply u-substitution

: \int \frac{1}{4 1 - u ^{2}} d u

= \int \frac{1}{4 1 - u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{1}{1 - u ^{2}} d u

Use the common integral:

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

= \frac{1}{4} arcsin (u)

Substitute back

u = t^{4}

= \frac{1}{4} arcsin (t^{4})

Add a constant to the solution

= \frac{1}{4} arcsin (t^{4}) + C

\int_{2}^{3} e^{1 - x} d x

\frac{\partial}{\partial y} (ln (y^{2} + 1))

\frac{d}{d x} (4 x^{2} - e^{x})

\frac{d y}{d x} = - \frac{1}{5 x y}

\int (\frac{ln ( x )}{x})^{2} d x

What is the integral of (t^3)/(sqrt(1-t^8)) ?
The integral of (t^3)/(sqrt(1-t^8)) is 1/4 arcsin(t^4)+C