What is the integral of sin^5(t) ?

The integral of sin^5(t) is -cos(t)+(2cos^3(t))/3-(cos^5(t))/5+C

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integral of sin^5(t)

\int sin^{5} (t) d t

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

Solution steps

\int sin^{5} (t) d t

Rewrite using trig identities

= \int (1 - cos^{2} (t))^{2} sin (t) d t

Apply u-substitution

: \int - 1 + 2 u^{2} - u^{4} d u

= \int - 1 + 2 u^{2} - u^{4} 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 1 d u + \int 2 u^{2} d u - \int u^{4} d u

\int 1 d u = u

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

\int u^{4} d u = \frac{u ^{5}}{5}

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

Substitute back

u = cos (t)

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

Add a constant to the solution

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

\int (x^{20} - 5 x^{4})^{6} (x^{19} - x^{3}) d x

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

f (x) = x^{2} + 5 x - 3, at x = - 3

\frac{\partial}{\partial y} (e^{1 + x^{2} - y^{2}})

\frac{d}{d x} (\frac{3}{x ^{4} + 2})

What is the integral of sin^5(t) ?
The integral of sin^5(t) is -cos(t)+(2cos^3(t))/3-(cos^5(t))/5+C