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

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

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integral of (x+1)cos(x^2+2x)

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

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

Solution steps

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

Apply u-substitution

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

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

Factor

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

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

Simplify

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{8 x ^{3}}{4 x ^{2} + 9} d x

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

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

\int tan (x + 1) d x

\int 2 x e^{x^{2} + 1} d x

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