What is the integral of tsqrt(1-t) ?

The integral of tsqrt(1-t) is 2/5 (1-t)^{5/2}-2/3 (1-t)^{3/2}+C

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integral of tsqrt(1-t)

\int t 1 - t d t

\frac{2}{5} (1 - t)^{\frac{5}{2}} - \frac{2}{3} (1 - t)^{\frac{3}{2}} + C

Solution steps

\int t 1 - t d t

Apply u-substitution

: \int - (- u + 1) u d u

= \int - (- u + 1) u d u

Take the constant out:

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

= - \int (- u + 1) u d u

Expand

(- u + 1) u : - u^{\frac{3}{2}} + u

= - \int - u^{\frac{3}{2}} + u d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- \int u^{\frac{3}{2}} d u + \int u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int u d u = \frac{2}{3} u^{\frac{3}{2}}

= - (- \frac{2}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}})

Substitute back

u = 1 - t

= - (- \frac{2}{5} (1 - t)^{\frac{5}{2}} + \frac{2}{3} (1 - t)^{\frac{3}{2}})

Simplify

= \frac{2}{5} (1 - t)^{\frac{5}{2}} - \frac{2}{3} (1 - t)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{5} (1 - t)^{\frac{5}{2}} - \frac{2}{3} (1 - t)^{\frac{3}{2}} + C

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What is the integral of tsqrt(1-t) ?
The integral of tsqrt(1-t) is 2/5 (1-t)^{5/2}-2/3 (1-t)^{3/2}+C