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

The integral of 2xcos(x^2-1) is -sin(1-x^2)+C

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integral of 2xcos(x^2-1)

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

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

Solution steps

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

Factor

x^{2} - 1 : - (- x^{2} + 1)

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

Simplify

x cos (- (- x^{2} + 1)) : x cos (1 - x^{2})

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

Apply u-substitution

: \int - \frac{cos ( u )}{2} d u

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

Take the constant out:

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

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

Use the common integral:

\int cos (u) d u = sin (u)

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

Substitute back

u = 1 - x^{2}

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

Simplify

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

= - sin (1 - x^{2})

Add a constant to the solution

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

\int \frac{1}{( x - 1 ) ^{2} \cdot ( x ^{2} + 3 )} d x

\int p (p + 4)^{7} d p

\int \frac{1}{u ^{8}} d u

\int x (x + 1) (x + 2) d x

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

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