What is the integral of sin^5(2t) ?

The integral of sin^5(2t) is 32(-(cos^6(t))/6+(cos^8(t))/4-(cos^{10}(t))/(10))+C

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integral of sin^5(2t)

\int sin^{5} (2 t) d t

32 (- \frac{cos ^{6} ( t )}{6} + \frac{cos ^{8} ( t )}{4} - \frac{cos ^{10} ( t )}{10}) + C

Solution steps

\int sin^{5} (2 t) d t

Rewrite using trig identities

= \int 32 cos^{5} (t) sin^{5} (t) d t

Take the constant out:

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

= 32 \cdot \int cos^{5} (t) sin^{5} (t) d t

Rewrite using trig identities

= 32 \cdot \int (1 - cos^{2} (t))^{2} sin (t) cos^{5} (t) d t

Apply u-substitution

: \int - u^{5} (1 - u^{2})^{2} d u

= 32 \cdot \int - u^{5} (1 - u^{2})^{2} d u

Expand

- u^{5} (1 - u^{2})^{2} : - u^{5} + 2 u^{7} - u^{9}

= 32 \cdot \int - u^{5} + 2 u^{7} - u^{9} 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

= 32 (- \int u^{5} d u + \int 2 u^{7} d u - \int u^{9} d u)

\int u^{5} d u = \frac{u ^{6}}{6}

\int 2 u^{7} d u = \frac{u ^{8}}{4}

\int u^{9} d u = \frac{u ^{10}}{10}

= 32 (- \frac{u ^{6}}{6} + \frac{u ^{8}}{4} - \frac{u ^{10}}{10})

Substitute back

u = cos (t)

= 32 (- \frac{cos ^{6} ( t )}{6} + \frac{cos ^{8} ( t )}{4} - \frac{cos ^{10} ( t )}{10})

Add a constant to the solution

= 32 (- \frac{cos ^{6} ( t )}{6} + \frac{cos ^{8} ( t )}{4} - \frac{cos ^{10} ( t )}{10}) + C

\frac{\partial}{\partial x} (3 sin (x cos (y)))

x \to 0 lim (x sin (\frac{1}{x}))

\int \frac{1}{1 - 2 x} d x

\frac{\partial}{\partial x} (x y (1 - x y))

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

What is the integral of sin^5(2t) ?
The integral of sin^5(2t) is 32(-(cos^6(t))/6+(cos^8(t))/4-(cos^{10}(t))/(10))+C