Question

integral f(x) - g(x) d(x) = f(x)X^3 - integral 3x^2 sinx d(x)

then f(x)-g(x)=.......

Answer 1

Answer on Question #37770 – Math - Integral Calculus

We have

$$\int (f(x) - g(x)) dx = f(x)x^3 - \int 3x^2 \sin x \, dx$$

Use this property of antiderivative to find $f(x) - g(x)$:

$$\int u dv = uv - \int v du$$

Then $udv = (f(x) - g(x))dx$, $vu = f(x)x^3$, $vdu = 3x^2 \sin x \, dx = \sin x \, dx^3$.

We can see that $v = f(x) = \sin x$, $u = x^3$.

Substitute it into (1):

$$\int (\sin x - g(x)) dx = x^3 \sin x - \int 3x^2 \sin x \, dx$$

Use this property to find $g(x)$:

$$\frac{d}{dx} \left( \int F(x) \, dx \right) = F(x)$$ $$\frac{d}{dx} (\dots) \left| \int (\sin x - g(x)) dx = x^3 \sin x - \int 3x^2 \sin x \, dx \right.$$ $$\begin{array}{l} \sin x - g(x) = x^3 \cos x + 3x^2 \sin x - 3x^2 \sin x \\ \sin x - x^3 \cos x = g(x) \end{array}$$

So

$$f(x) - g(x) = \sin x - \sin x + x^3 \cos x = x^3 \cos x$$

Answer: $f(x) - g(x) = x^3 \cos x$.