What is the integral of 2xcos(sqrt(2x^2+3)) ?

The integral of 2xcos(sqrt(2x^2+3)) is sin(sqrt(2x^2+3))sqrt(2x^2+3)+cos(sqrt(2x^2+3))+C

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integral of 2xcos(sqrt(2x^2+3))

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

sin (2 x^{2} + 3) 2 x^{2} + 3 + cos (2 x^{2} + 3) + C

Solution steps

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

Apply u-substitution

: \int \frac{cos ( u )}{4} d u

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

Take the constant out:

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

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

Apply u-substitution

: \int cos (v) \cdot 2 v d v

= 2 \cdot \frac{1}{4} \cdot \int cos (v) \cdot 2 v d v

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot 2 \cdot \int cos (v) v d v

Apply Integration By Parts

: v sin (v) - \int sin (v) d v

= 2 \cdot \frac{1}{4} \cdot 2 (v sin (v) - \int sin (v) d v)

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

= 2 \cdot \frac{1}{4} \cdot 2 (v sin (v) - (- cos (v)))

Substitute back

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

Simplify

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

= sin (2 x^{2} + 3) 2 x^{2} + 3 + cos (2 x^{2} + 3)

Add a constant to the solution

= sin (2 x^{2} + 3) 2 x^{2} + 3 + cos (2 x^{2} + 3) + C

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

\int 4 x^{3} x^{4} - 7 d x

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

What is the integral of 2xcos(sqrt(2x^2+3)) ?
The integral of 2xcos(sqrt(2x^2+3)) is sin(sqrt(2x^2+3))sqrt(2x^2+3)+cos(sqrt(2x^2+3))+C