What is the integral of sin(x^2-5)x ?

The integral of sin(x^2-5)x is 1/2 (x^2sin(x^2-5)+x^2sin(5-x^2)-cos(5-x^2))+C

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integral of sin(x^2-5)x

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

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

Solution steps

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

Apply u-substitution

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

= \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

= \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

= \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)

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

Substitute back

u = x^{2}

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

Expand

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

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

Add a constant to the solution

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

y^{^{'}} = x^{\frac{2}{3}}

\int_{1}^{e^{4}} \frac{ln ( x )}{x ^{2}} d x

x \to in f ini t o lim (10)

\frac{\partial}{\partial x} (5 (y^{2} - x^{2}) ln (x + y))

t^{2} y^{^{''}} - 2 y = 0

What is the integral of sin(x^2-5)x ?
The integral of sin(x^2-5)x is 1/2 (x^2sin(x^2-5)+x^2sin(5-x^2)-cos(5-x^2))+C