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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Use the common integral:

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

= tan (u)

Substitute back

u = arcsin (t)

= tan (arcsin (t))

Use the following identity:

tan (arcsin (x)) = \frac{x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

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

Add a constant to the solution

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

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