What is the integral of 2xsin^4(x^2-1) ?

The integral of 2xsin^4(x^2-1) is -1/4 sin^3(x^2-1)cos(x^2-1)+3/8 (x^2-1-1/2 sin(2(x^2-1)))+C

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integral of 2xsin^4(x^2-1)

\int 2 x sin^{4} (x^{2} - 1) d x

- \frac{1}{4} sin^{3} (x^{2} - 1) cos (x^{2} - 1) + \frac{3}{8} (x^{2} - 1 - \frac{1}{2} sin (2 (x^{2} - 1))) + C

Solution steps

\int 2 x sin^{4} (x^{2} - 1) d x

Take the constant out:

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

= 2 \cdot \int x sin^{4} (x^{2} - 1) d x

Apply u-substitution

: \int \frac{sin ^{4} ( u )}{2} d u

= 2 \cdot \int \frac{sin ^{4} ( u )}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int sin^{4} (u) d u

Apply Integral Reduction

= 2 \cdot \frac{1}{2} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \int sin^{2} (u) d u)

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

= 2 \cdot \frac{1}{2} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u)))

Substitute back

u = x^{2} - 1

= 2 \cdot \frac{1}{2} (- \frac{cos ( x ^{2} - 1 ) sin ^{3} ( x ^{2} - 1 )}{4} + \frac{3}{4} \cdot \frac{1}{2} (x^{2} - 1 - \frac{1}{2} sin (2 (x^{2} - 1))))

Simplify

2 \cdot \frac{1}{2} (- \frac{cos ( x ^{2} - 1 ) sin ^{3} ( x ^{2} - 1 )}{4} + \frac{3}{4} \cdot \frac{1}{2} (x^{2} - 1 - \frac{1}{2} sin (2 (x^{2} - 1)))) : - \frac{1}{4} sin^{3} (x^{2} - 1) cos (x^{2} - 1) + \frac{3}{8} (x^{2} - 1 - \frac{1}{2} sin (2 (x^{2} - 1)))

= - \frac{1}{4} sin^{3} (x^{2} - 1) cos (x^{2} - 1) + \frac{3}{8} (x^{2} - 1 - \frac{1}{2} sin (2 (x^{2} - 1)))

Add a constant to the solution

= - \frac{1}{4} sin^{3} (x^{2} - 1) cos (x^{2} - 1) + \frac{3}{8} (x^{2} - 1 - \frac{1}{2} sin (2 (x^{2} - 1))) + C

x \to 0 lim (2 x^{2} + 5)

\int (x^{1.3} + 11 x^{4.5}) d x

1 + 2 t

\int \frac{u}{u ^{2} + 1} d u

\int \frac{ln ( x ) + 1}{x} d x

What is the integral of 2xsin^4(x^2-1) ?
The integral of 2xsin^4(x^2-1) is -1/4 sin^3(x^2-1)cos(x^2-1)+3/8 (x^2-1-1/2 sin(2(x^2-1)))+C