What is the integral of (4sin(x))^3 ?

The integral of (4sin(x))^3 is 64(-cos(x)+(cos^3(x))/3)+C

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integral of (4sin(x))^3

\int (4 sin (x))^{3} d x

64 (- cos (x) + \frac{cos ^{3} ( x )}{3}) + C

Solution steps

\int (4 sin (x))^{3} d x

Simplify

(4 sin (x))^{3} : 64 sin^{3} (x)

= \int 64 sin^{3} (x) d x

Take the constant out:

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

= 64 \cdot \int sin^{3} (x) d x

Rewrite using trig identities

= 64 \cdot \int (1 - cos^{2} (x)) sin (x) d x

Apply u-substitution

: \int - 1 + u^{2} d u

= 64 \cdot \int - 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

= 64 (- \int 1 d u + \int u^{2} d u)

\int 1 d u = u

\int u^{2} d u = \frac{u ^{3}}{3}

= 64 (- u + \frac{u ^{3}}{3})

Substitute back

u = cos (x)

= 64 (- cos (x) + \frac{cos ^{3} ( x )}{3})

Add a constant to the solution

= 64 (- cos (x) + \frac{cos ^{3} ( x )}{3}) + C

\frac{3 ( s + 1 )}{s ^{2} + \frac{1}{2}}

e^{- \frac{x}{2}}, 0

n = 2 \sum \infty (\frac{1}{2})^{3 n}

\int \frac{1}{( - u ^{2} + u - 1 )} d u

\frac{d}{d x} (e^{c o s (x)} + cos (e^{x}))

What is the integral of (4sin(x))^3 ?
The integral of (4sin(x))^3 is 64(-cos(x)+(cos^3(x))/3)+C