What is the integral of (y^2)/((49-y^2)^{3/2)} ?

The integral of (y^2)/((49-y^2)^{3/2)} is -arcsin(1/7 y)+y/((49-y^2)^{1/2)}+C

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integral of (y^2)/((49-y^2)^{3/2)}

\int \frac{y ^{2}}{( 49 - y ^{2} ) ^{\frac{3}{2}}} d y

- arcsin (\frac{1}{7} y) + \frac{y}{( 49 - y ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{y ^{2}}{( 49 - y ^{2} ) ^{\frac{3}{2}}} d y

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}{7} y)

= - arcsin (\frac{1}{7} y) + tan (arcsin (\frac{1}{7} y))

Simplify

- arcsin (\frac{1}{7} y) + tan (arcsin (\frac{1}{7} y)) : - arcsin (\frac{1}{7} y) + \frac{y}{( 49 - y ^{2} ) ^{\frac{1}{2}}}

= - arcsin (\frac{1}{7} y) + \frac{y}{( 49 - y ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= - arcsin (\frac{1}{7} y) + \frac{y}{( 49 - y ^{2} ) ^{\frac{1}{2}}} + C

\int (3 x - 6) e^{x^{2} - 4 x} d x

cos^{2} (x) sin^{2} (x)

n = 1 \sum \infty \frac{n !}{10 5 ^{n}}

\frac{\partial}{\partial x} (x^{2} - 3 x y)

x 1 - y^{2} d x = d y

What is the integral of (y^2)/((49-y^2)^{3/2)} ?
The integral of (y^2)/((49-y^2)^{3/2)} is -arcsin(1/7 y)+y/((49-y^2)^{1/2)}+C