What is the integral of t/(sqrt(1-t^2)) ?

The integral of t/(sqrt(1-t^2)) is -sqrt(1-t^2)+C

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integral of t/(sqrt(1-t^2))

\int \frac{t}{1 - t ^{2}} d t

- 1 - t^{2} + C

Solution steps

\int \frac{t}{1 - t ^{2}} d t

Apply u-substitution

: \int - \frac{1}{2 u} d u

= \int - \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{2} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= - \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = 1 - t^{2}

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

Simplify

- \frac{1}{2} \cdot 2 (1 - t^{2})^{\frac{1}{2}} : - 1 - t^{2}

= - 1 - t^{2}

Add a constant to the solution

= - 1 - t^{2} + C

sin (x) d x + y d y = 0, y (0) = - 2

\int \frac{1}{( x ^{2} + 10 x + 24 )} d x

\frac{d}{d x} (ln (arctan (x)))

\frac{d y}{d x} - 3 y = 0

n = 1 \sum \infty (- 0.7)^{n - 1}

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