What is the integral of 2x(cos(x^2-5))/(cot(x^2-5)) ?

The integral of 2x(cos(x^2-5))/(cot(x^2-5)) is x^2sin(x^2-5)+x^2sin(5-x^2)-cos(5-x^2)+C

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integral of 2x(cos(x^2-5))/(cot(x^2-5))

\int 2 x \frac{cos ( x ^{2} - 5 )}{cot ( x ^{2} - 5 )} d x

x^{2} sin (x^{2} - 5) + x^{2} sin (5 - x^{2}) - cos (5 - x^{2}) + C

Solution steps

\int 2 x \frac{cos ( x ^{2} - 5 )}{cot ( x ^{2} - 5 )} d x

Take the constant out:

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

= 2 \cdot \int x \frac{cos ( x ^{2} - 5 )}{cot ( x ^{2} - 5 )} d x

Rewrite using trig identities

= 2 \cdot \int x sin (x^{2} - 5) d x

Apply u-substitution

: \int \frac{sin ( u - 5 )}{2} d u

= 2 \cdot \int \frac{sin ( u - 5 )}{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 (u - 5) d u

Apply Integration By Parts

: sin (u - 5) u - \int cos (u - 5) u d u

= 2 \cdot \frac{1}{2} (sin (u - 5) u - \int cos (u - 5) u d u)

\int cos (u - 5) u d u = - u sin (5 - u) + cos (5 - u)

= 2 \cdot \frac{1}{2} (sin (u - 5) u - (- u sin (5 - u) + cos (5 - u)))

Substitute back

u = x^{2}

= 2 \cdot \frac{1}{2} (sin (x^{2} - 5) x^{2} - (- x^{2} sin (5 - x^{2}) + cos (5 - x^{2})))

Simplify

2 \cdot \frac{1}{2} (sin (x^{2} - 5) x^{2} - (- x^{2} sin (5 - x^{2}) + cos (5 - x^{2}))) : x^{2} sin (x^{2} - 5) + x^{2} sin (5 - x^{2}) - cos (5 - x^{2})

= x^{2} sin (x^{2} - 5) + x^{2} sin (5 - x^{2}) - cos (5 - x^{2})

Add a constant to the solution

= x^{2} sin (x^{2} - 5) + x^{2} sin (5 - x^{2}) - cos (5 - x^{2}) + C

\int e^{2 x} \cdot 4 x d x

\int \frac{x - 2}{x ^{2} + 6 x + 13} d x

\int \frac{2 x ^{2} - 1}{1 + x ^{2}} d x

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

\int \frac{x ^{2} - 5}{x ^{2}} d x

What is the integral of 2x(cos(x^2-5))/(cot(x^2-5)) ?
The integral of 2x(cos(x^2-5))/(cot(x^2-5)) is x^2sin(x^2-5)+x^2sin(5-x^2)-cos(5-x^2)+C