What is the integral of (16x^3)/(sqrt(25-16x^8)) ?

The integral of (16x^3)/(sqrt(25-16x^8)) is arcsin(4/5 x^4)+C

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integral of (16x^3)/(sqrt(25-16x^8))

\int \frac{16 x ^{3}}{25 - 16 x ^{8}} d x

arcsin (\frac{4}{5} x^{4}) + C

Solution steps

\int \frac{16 x ^{3}}{25 - 16 x ^{8}} d x

Take the constant out:

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

= 16 \cdot \int \frac{x ^{3}}{25 - 16 x ^{8}} d x

Apply u-substitution

= 16 \cdot \int \frac{1}{4 25 - 16 u ^{2}} d u

Take the constant out:

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

= 16 \cdot \frac{1}{4} \cdot \int \frac{1}{25 - 16 u ^{2}} d u

Apply Trigonometric Substitution

= 16 \cdot \frac{1}{4} \cdot \int \frac{1}{4} d v

Take the constant out:

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

= 16 \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 16 \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot 1 \cdot v

Substitute back

= 16 \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{5} x^{4})

Simplify

16 \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{5} x^{4}) : arcsin (\frac{4}{5} x^{4})

= arcsin (\frac{4}{5} x^{4})

Add a constant to the solution

= arcsin (\frac{4}{5} x^{4}) + C

What is the integral of (16x^3)/(sqrt(25-16x^8)) ?
The integral of (16x^3)/(sqrt(25-16x^8)) is arcsin(4/5 x^4)+C