What is the integral of 1/((7-t^2)^{3/2)} ?

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

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integral of 1/((7-t^2)^{3/2)}

\int \frac{1}{( 7 - t ^{2} ) ^{\frac{3}{2}}} d t

\frac{t}{7 ( 7 - t ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 7 - t ^{2} ) ^{\frac{3}{2}}} d t

Apply Trigonometric Substitution

: \int \frac{1}{7 cos ^{2} ( u )} d u

= \int \frac{1}{7 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}{7} \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}{7} tan (u)

Substitute back

u = arcsin (\frac{1}{7 ^{\frac{1}{2}}} t)

= \frac{1}{7} tan (arcsin (\frac{1}{7 ^{\frac{1}{2}}} t))

Simplify

\frac{1}{7} tan (arcsin (\frac{1}{7 ^{\frac{1}{2}}} t)) : \frac{t}{7 ( 7 - t ^{2} ) ^{\frac{1}{2}}}

= \frac{t}{7 ( 7 - t ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{t}{7 ( 7 - t ^{2} ) ^{\frac{1}{2}}} + C

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