What is the integral of 1/((16-t^2)^{3/2)} ?

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

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integral of 1/((16-t^2)^{3/2)}

\int \frac{1}{( 16 - t ^{2} ) ^{\frac{3}{2}}} d t

\frac{t}{16 ( 16 - t ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 16 - t ^{2} ) ^{\frac{3}{2}}} d t

Apply Trigonometric Substitution

: \int \frac{1}{16 cos ^{2} ( u )} d u

= \int \frac{1}{16 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}{16} \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}{16} tan (u)

Substitute back

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

= \frac{1}{16} tan (arcsin (\frac{1}{4} t))

Simplify

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

= \frac{t}{16 ( 16 - t ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{t}{16 ( 16 - t ^{2} ) ^{\frac{1}{2}}} + C

\int_{- 1}^{3} 30 - 3 y^{2} - \frac{9}{2} d y

\frac{d y}{d x} = \frac{1}{1 + x ^{2}}

x \to 0 lim (\frac{5}{1 - e ^{\frac{1}{x}}})

16 y^{^{''}} + 11 y = 0

\int \frac{e ^{2 x}}{e ^{x} - 1} d x

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