What is the integral from 0 to pi/4 of cos^7(x) ?

The integral from 0 to pi/4 of cos^7(x) is (177)/(280sqrt(2))

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integral from 0 to pi/4 of cos^7(x)

\int_{0}^{\frac{π}{4}} cos^{7} (x) d x

\frac{177}{280 2}

Decimal Notation

0.44699 \dots

Solution steps

\int_{0}^{\frac{π}{4}} cos^{7} (x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{4}} (1 - sin^{2} (x))^{3} cos (x) d x

Apply u-substitution

: \int_{0}^{\frac{2}{2}} (1 - u^{2})^{3} d u

= \int_{0}^{\frac{2}{2}} (1 - u^{2})^{3} d u

Expand

(1 - u^{2})^{3} : 1 - 3 u^{2} + 3 u^{4} - u^{6}

= \int_{0}^{\frac{2}{2}} 1 - 3 u^{2} + 3 u^{4} - u^{6} 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_{0}^{\frac{2}{2}} 1 d u - \int_{0}^{\frac{2}{2}} 3 u^{2} d u + \int_{0}^{\frac{2}{2}} 3 u^{4} d u - \int_{0}^{\frac{2}{2}} u^{6} d u

\int_{0}^{\frac{2}{2}} 1 d u = \frac{1}{2}

\int_{0}^{\frac{2}{2}} 3 u^{2} d u = \frac{1}{2 2}

\int_{0}^{\frac{2}{2}} 3 u^{4} d u = \frac{3}{20 2}

\int_{0}^{\frac{2}{2}} u^{6} d u = \frac{1}{56 2}

= \frac{1}{2} - \frac{1}{2 2} + \frac{3}{20 2} - \frac{1}{56 2}

Simplify

\frac{1}{2} - \frac{1}{2 2} + \frac{3}{20 2} - \frac{1}{56 2} : \frac{177}{280 2}

= \frac{177}{280 2}

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

\frac{x ^{3}}{8}

\int cos^{4} (6 x) d x

\frac{d}{d x} (\frac{1}{e} e^{x})

\frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x - 2}

What is the integral from 0 to pi/4 of cos^7(x) ?
The integral from 0 to pi/4 of cos^7(x) is (177)/(280sqrt(2))