What is the integral of (x^2)/((16-x^2)^{3/2)} ?

The integral of (x^2)/((16-x^2)^{3/2)} is -arcsin(1/4 x)+x/((16-x^2)^{1/2)}+C

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integral of (x^2)/((16-x^2)^{3/2)}

\int \frac{x ^{2}}{( 16 - x ^{2} ) ^{\frac{3}{2}}} d x

- arcsin (\frac{1}{4} x) + \frac{x}{( 16 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{x ^{2}}{( 16 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int tan^{2} (u) d u

= \int tan^{2} (u) d u

Rewrite using trig identities

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

Apply the Sum Rule:

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

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

\int 1 d u = u

\int sec^{2} (u) d u = tan (u)

= - u + tan (u)

Substitute back

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

= - arcsin (\frac{1}{4} x) + tan (arcsin (\frac{1}{4} x))

Simplify

- arcsin (\frac{1}{4} x) + tan (arcsin (\frac{1}{4} x)) : - arcsin (\frac{1}{4} x) + \frac{x}{( 16 - x ^{2} ) ^{\frac{1}{2}}}

= - arcsin (\frac{1}{4} x) + \frac{x}{( 16 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= - arcsin (\frac{1}{4} x) + \frac{x}{( 16 - x ^{2} ) ^{\frac{1}{2}}} + C

\frac{d}{d x} (e^{3 x} \cdot 3)

\int cos (6 y) - cos (- 6 y) d y

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What is the integral of (x^2)/((16-x^2)^{3/2)} ?
The integral of (x^2)/((16-x^2)^{3/2)} is -arcsin(1/4 x)+x/((16-x^2)^{1/2)}+C