What is the integral of sin^3(2x+1) ?

The integral of sin^3(2x+1) is 1/2 (-cos(2x+1)+(cos^3(2x+1))/3)+C

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integral of sin^3(2x+1)

\int sin^{3} (2 x + 1) d x

\frac{1}{2} (- cos (2 x + 1) + \frac{cos ^{3} ( 2 x + 1 )}{3}) + C

Solution steps

\int sin^{3} (2 x + 1) d x

Apply u-substitution

: \int sin^{3} (u) \frac{1}{2} d u

= \int sin^{3} (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int sin^{3} (u) d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int (1 - cos^{2} (u)) sin (u) d u

Apply u-substitution

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

= \frac{1}{2} \cdot \int - 1 + v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d v = v

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

= \frac{1}{2} (- v + \frac{v ^{3}}{3})

Substitute back

= \frac{1}{2} (- cos (2 x + 1) + \frac{cos ^{3} ( 2 x + 1 )}{3})

Add a constant to the solution

= \frac{1}{2} (- cos (2 x + 1) + \frac{cos ^{3} ( 2 x + 1 )}{3}) + C

\int_{0}^{0.5} 12 (1 + 2 x^{2}) d x

\int 19 x d x

f (x) = \frac{( x - 3 ) ^{2}}{( x + 9 )}

8 x + 8 x^{- 1}

\frac{d}{d x} (e^{- \frac{x}{y}})

What is the integral of sin^3(2x+1) ?
The integral of sin^3(2x+1) is 1/2 (-cos(2x+1)+(cos^3(2x+1))/3)+C