What is the integral from 0 to pi/4 of cos^3(θ) ?

The integral from 0 to pi/4 of cos^3(θ) is 5/(6sqrt(2))

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integral from 0 to pi/4 of cos^3(θ)

\int_{0}^{\frac{π}{4}} cos^{3} (θ) d θ

\frac{5}{6 2}

Decimal

0.58925 \dots

Solution steps

\int_{0}^{\frac{π}{4}} cos^{3} (θ) d θ

Rewrite using trig identities

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

Apply u-substitution

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

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

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

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

= \frac{1}{2} - \frac{1}{6 2}

Simplify

\frac{1}{2} - \frac{1}{6 2} : \frac{5}{6 2}

= \frac{5}{6 2}

\int_{1}^{5} (x^{2} + 4 x + 5) d x

\int_{1}^{5} - x^{2} + 6 x - 5 d x

\int_{0}^{2 π} sin^{3} (t) d t

\int_{1}^{2} \frac{7}{x - 2} d x

\int_{3}^{15} \frac{5 x ^{2} - 9}{3} d x

What is the integral from 0 to pi/4 of cos^3(θ) ?
The integral from 0 to pi/4 of cos^3(θ) is 5/(6sqrt(2))