What is the integral of (t^5)/(sqrt(1-t^{12))} ?

The integral of (t^5)/(sqrt(1-t^{12))} is 1/6 arcsin(t^6)+C

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integral of (t^5)/(sqrt(1-t^{12))}

\int \frac{t ^{5}}{1 - t ^{12}} d t

\frac{1}{6} arcsin (t^{6}) + C

Solution steps

\int \frac{t ^{5}}{1 - t ^{12}} d t

Apply u-substitution

: \int \frac{1}{6 1 - u ^{2}} d u

= \int \frac{1}{6 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}{6} \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}{6} arcsin (u)

Substitute back

u = t^{6}

= \frac{1}{6} arcsin (t^{6})

Add a constant to the solution

= \frac{1}{6} arcsin (t^{6}) + C

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\int 375 - 25 15 x d x

cos (x)

What is the integral of (t^5)/(sqrt(1-t^{12))} ?
The integral of (t^5)/(sqrt(1-t^{12))} is 1/6 arcsin(t^6)+C