What is the integral of 1/((6-y^2)^{3/2)} ?

The integral of 1/((6-y^2)^{3/2)} is y/(6(6-y^2)^{1/2)}+C

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integral of 1/((6-y^2)^{3/2)}

\int \frac{1}{( 6 - y ^{2} ) ^{\frac{3}{2}}} d y

\frac{y}{6 ( 6 - y ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 6 - y ^{2} ) ^{\frac{3}{2}}} d y

Apply Trigonometric Substitution

: \int \frac{1}{6 cos ^{2} ( u )} d u

= \int \frac{1}{6 cos ^{2} ( u )} 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}{cos ^{2} ( u )} d u

Use the common integral:

\int \frac{1}{cos ^{2} ( u )} d u = tan (u)

= \frac{1}{6} tan (u)

Substitute back

u = arcsin (\frac{1}{6 ^{\frac{1}{2}}} y)

= \frac{1}{6} tan (arcsin (\frac{1}{6 ^{\frac{1}{2}}} y))

Simplify

\frac{1}{6} tan (arcsin (\frac{1}{6 ^{\frac{1}{2}}} y)) : \frac{y}{6 ( 6 - y ^{2} ) ^{\frac{1}{2}}}

= \frac{y}{6 ( 6 - y ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{y}{6 ( 6 - y ^{2} ) ^{\frac{1}{2}}} + C

x^{3} y^{^{'''}} - x^{2} y^{^{''}} + 2 x y^{^{'}} - 2 y = x^{3}

\int \frac{1}{x ^{2} + \frac{3}{4}} d x

\frac{1}{x ^{2} + 3}

f (x) = - x^{3}, (- 1, 1)

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

What is the integral of 1/((6-y^2)^{3/2)} ?
The integral of 1/((6-y^2)^{3/2)} is y/(6(6-y^2)^{1/2)}+C