What is the integral from 0 to pi/2 of 9sin^3(θ) ?

The integral from 0 to pi/2 of 9sin^3(θ) is 6

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integral from 0 to pi/2 of 9sin^3(θ)

\int_{0}^{\frac{π}{2}} 9 sin^{3} (θ) d θ

6

Solution steps

\int_{0}^{\frac{π}{2}} 9 sin^{3} (θ) d θ

Take the constant out:

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

= 9 \cdot \int_{0}^{\frac{π}{2}} sin^{3} (θ) d θ

Rewrite using trig identities

= 9 \cdot \int_{0}^{\frac{π}{2}} (1 - cos^{2} (θ)) sin (θ) d θ

Apply u-substitution

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

= 9 \cdot \int_{1}^{0} - 1 + u^{2} d u

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

= 9 (- \int_{0}^{1} - 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

= 9 (- (- \int_{0}^{1} 1 d u + \int_{0}^{1} u^{2} d u))

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

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

= 9 (- (- 1 + \frac{1}{3}))

Simplify

9 (- (- 1 + \frac{1}{3})) : 6

= 6

\int_{- 2}^{x} (8 t^{2} - 7 t) d t

\int_{0}^{\infty} e^{- 3} d x

\int_{- 1}^{3} x (x^{2} - 3) d x

\int_{1}^{2} (8 t^{3} - 6 t^{- 2}) d t

\int_{0}^{π} 1 - cos (θ) d θ

What is the integral from 0 to pi/2 of 9sin^3(θ) ?
The integral from 0 to pi/2 of 9sin^3(θ) is 6