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

The integral from 0 to pi/2 of 3(sin(x))^3 is 2

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integral from 0 to pi/2 of 3(sin(x))^3

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

2

Solution steps

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

Take the constant out:

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

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

Simplify

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

Rewrite using trig identities

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

Apply u-substitution

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

= 3 \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

= 3 (- \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

= 3 (- (- \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}

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

Simplify

3 (- (- 1 + \frac{1}{3})) : 2

= 2

\int_{0}^{2 π} sin (y) d y

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\int_{- 1}^{3} (6 x^{3} + 3 x^{2} - 7 x) d x

\int_{0}^{6} (x^{2} + 3) d x

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