What is the integral of 7xcos(x^2-3) ?

The integral of 7xcos(x^2-3) is 7/2 sin(x^2-3)+C

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integral of 7xcos(x^2-3)

\int 7 x cos (x^{2} - 3) d x

\frac{7}{2} sin (x^{2} - 3) + C

Solution steps

\int 7 x cos (x^{2} - 3) d x

Take the constant out:

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

= 7 \cdot \int x cos (x^{2} - 3) d x

Apply u-substitution

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

= 7 \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

= 7 \cdot \frac{1}{2} \cdot \int cos (u) d u

Use the common integral:

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

= 7 \cdot \frac{1}{2} sin (u)

Substitute back

u = x^{2} - 3

= 7 \cdot \frac{1}{2} sin (x^{2} - 3)

Simplify

7 \cdot \frac{1}{2} sin (x^{2} - 3) : \frac{7}{2} sin (x^{2} - 3)

= \frac{7}{2} sin (x^{2} - 3)

Add a constant to the solution

= \frac{7}{2} sin (x^{2} - 3) + C

\int \frac{1 + 4 cot ( x )}{4 - cot ( x )} d x

\int cot^{3} (d) x d x

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What is the integral of 7xcos(x^2-3) ?
The integral of 7xcos(x^2-3) is 7/2 sin(x^2-3)+C