What is the integral of 5xsqrt(1-x^4) ?

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

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integral of 5xsqrt(1-x^4)

\int 5 x 1 - x^{4} d x

\frac{5}{4} (arcsin (x^{2}) + \frac{1}{2} sin (2 arcsin (x^{2}))) + C

Solution steps

\int 5 x 1 - x^{4} d x

Take the constant out:

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

= 5 \cdot \int x 1 - x^{4} d x

Simplify

x^{4} : (x^{2})^{2}

= 5 \cdot \int x 1 - (x^{2})^{2} d x

Apply u-substitution

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

= 5 \cdot \int \frac{1 - u ^{2}}{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}{2} \cdot \int 1 - u^{2} d u

Apply Trigonometric Substitution

: \int cos^{2} (v) d v

= 5 \cdot \frac{1}{2} \cdot \int cos^{2} (v) d v

Rewrite using trig identities

= 5 \cdot \frac{1}{2} \cdot \int \frac{1 + cos ( 2 v )}{2} d v

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \int 1 + cos (2 v) 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

= 5 \cdot \frac{1}{2} \cdot \frac{1}{2} (\int 1 d v + \int cos (2 v) d v)

\int 1 d v = v

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

= 5 \cdot \frac{1}{2} \cdot \frac{1}{2} (v + \frac{1}{2} sin (2 v))

Substitute back

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

Simplify

5 \cdot \frac{1}{2} \cdot \frac{1}{2} (arcsin (x^{2}) + \frac{1}{2} sin (2 arcsin (x^{2}))) : \frac{5}{4} (arcsin (x^{2}) + \frac{1}{2} sin (2 arcsin (x^{2})))

= \frac{5}{4} (arcsin (x^{2}) + \frac{1}{2} sin (2 arcsin (x^{2})))

Add a constant to the solution

= \frac{5}{4} (arcsin (x^{2}) + \frac{1}{2} sin (2 arcsin (x^{2}))) + C

x (x^{2} + 4)^{5}

f (x) = 5 x + 2 x^{2} + 7 e^{x}

y = - x^{2} + 1, y = 0

\int_{0}^{π} \frac{1}{π - x} d x

x \to - \frac{5}{2} lim (- x + 2)

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