What is the integral from 0 to pi/4 of sin^5(2x) ?

The integral from 0 to pi/4 of sin^5(2x) is 4/15

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integral from 0 to pi/4 of sin^5(2x)

\int_{0}^{\frac{π}{4}} sin^{5} (2 x) d x

\frac{4}{15}

Decimal Notation

0.26666 \dots

Solution steps

\int_{0}^{\frac{π}{4}} sin^{5} (2 x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{4}} 32 cos^{5} (x) sin^{5} (x) d x

Take the constant out:

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

= 32 \cdot \int_{0}^{\frac{π}{4}} cos^{5} (x) sin^{5} (x) d x

Rewrite using trig identities

= 32 \cdot \int_{0}^{\frac{π}{4}} (1 - cos^{2} (x))^{2} sin (x) cos^{5} (x) d x

Apply u-substitution

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

= 32 \cdot \int_{1}^{\frac{2}{2}} - 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_{1}^{\frac{2}{2}} - u^{5} + 2 u^{7} - u^{9} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 32 (- \int_{\frac{2}{2}}^{1} - 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_{\frac{2}{2}}^{1} u^{5} d u + \int_{\frac{2}{2}}^{1} 2 u^{7} d u - \int_{\frac{2}{2}}^{1} u^{9} d u))

\int_{\frac{2}{2}}^{1} u^{5} d u = \frac{7}{48}

\int_{\frac{2}{2}}^{1} 2 u^{7} d u = \frac{15}{64}

\int_{\frac{2}{2}}^{1} u^{9} d u = \frac{31}{320}

= 32 (- (- \frac{7}{48} + \frac{15}{64} - \frac{31}{320}))

Simplify

32 (- (- \frac{7}{48} + \frac{15}{64} - \frac{31}{320})) : \frac{4}{15}

= \frac{4}{15}

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

x \to \infty lim \frac{( ( a ^{(\frac{1}{x})} + b ^{(\frac{1}{x})} ) ^{x} )}{2 ^{x}}

y^{^{'}} + 10 (tan (10 x)) y = 9 cos (10 x)

\int (3 x^{2} + 4) x - 1 d x

\frac{d x}{d t} = - \frac{1}{2} x + t

What is the integral from 0 to pi/4 of sin^5(2x) ?
The integral from 0 to pi/4 of sin^5(2x) is 4/15