What is the integral of 5x^4sin^2(x^5) ?

The integral of 5x^4sin^2(x^5) is 1/2 (x^5-1/2 sin(2x^5))+C

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integral of 5x^4sin^2(x^5)

\int 5 x^{4} sin^{2} (x^{5}) d x

\frac{1}{2} (x^{5} - \frac{1}{2} sin (2 x^{5})) + C

Solution steps

\int 5 x^{4} sin^{2} (x^{5}) d x

Take the constant out:

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

= 5 \cdot \int x^{4} sin^{2} (x^{5}) d x

Apply u-substitution

: \int \frac{sin ^{2} ( u )}{5} d u

= 5 \cdot \int \frac{sin ^{2} ( u )}{5} d u

Take the constant out:

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

= 5 \cdot \frac{1}{5} \cdot \int sin^{2} (u) d u

Rewrite using trig identities

= 5 \cdot \frac{1}{5} \cdot \int \frac{1 - cos ( 2 u )}{2} d u

Take the constant out:

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

= 5 \cdot \frac{1}{5} \cdot \frac{1}{2} \cdot \int 1 - cos (2 u) 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

= 5 \cdot \frac{1}{5} \cdot \frac{1}{2} (\int 1 d u - \int cos (2 u) d u)

\int 1 d u = u

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

= 5 \cdot \frac{1}{5} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = x^{5}

= 5 \cdot \frac{1}{5} \cdot \frac{1}{2} (x^{5} - \frac{1}{2} sin (2 x^{5}))

Simplify

5 \cdot \frac{1}{5} \cdot \frac{1}{2} (x^{5} - \frac{1}{2} sin (2 x^{5})) : \frac{1}{2} (x^{5} - \frac{1}{2} sin (2 x^{5}))

= \frac{1}{2} (x^{5} - \frac{1}{2} sin (2 x^{5}))

Add a constant to the solution

= \frac{1}{2} (x^{5} - \frac{1}{2} sin (2 x^{5})) + C

\frac{d y}{d x} = e^{2 x - 5 y}

y^{^{'}} = \frac{x ^{2} + 2 y ^{2}}{x y}

y^{^{''}} - y^{^{'}} - 30 = 0

569 + t (37 t^{0.6} - 108)

\frac{d r}{d θ} = \frac{r ^{2} + 10 r θ}{5 θ ^{2}}

What is the integral of 5x^4sin^2(x^5) ?
The integral of 5x^4sin^2(x^5) is 1/2 (x^5-1/2 sin(2x^5))+C