Answer :
Main Answer :
[tex]The integral of \(x\arcsin(6x^2)\) with respect to \(x\) is \(\frac{1}{12}(6x^2\arcsin(6x^2) + \sqrt{1 - 36x^4}) + C\), where \(C\) is the constant of integration.[/tex]
Explanation:
[tex]To solve the integral, we first use the substitution method. Let \(u = 6x^2\), then \(du = 12x \, dx\). We can rewrite the integral as \(\frac{1}{12}\int x \cdot \arcsin(u) \, du\). Integrating by parts, we let \(dv = \arcsin(u) \, du\) and \(u = x \, dx\). This gives us \(v = u\arcsin(u) - \int \frac{u}{\sqrt{1-u^2}} \, du\).[/tex]
[tex]The remaining integral, \(\int \frac{u}{\sqrt{1-u^2}} \, du\), can be solved with the substitution \(w = 1 - u^2\), which gives \(dw = -2u \, du\). Thus, the integral becomes \(-\frac{1}{2}\int \frac{1}{\sqrt{w}} \, dw = -\sqrt{w} = -\sqrt{1 - u^2}\).[/tex]
[tex]Putting everything together, we have \(\frac{1}{12}\left(u\arcsin(u) + \sqrt{1 - u^2}\right) + C\). Substituting back \(u = 6x^2\), we get the final result: \(\frac{1}{12}\left(6x^2\arcsin(6x^2) + \sqrt{1 - 36x^4}\right) + C\).[/tex]
The derivative of the function [tex]\(g(x) = -3x^5(-7x^5)\)[/tex] can be calculated using the Product Rule, resulting in [tex]\(g'(x) = -105x^5 + 35x^{10}\)[/tex]. Hence the correct option is c) [tex]-105x^5 + 35x^{10}[/tex]
To find [tex]\(g'(x)\)[/tex] using the Product Rule, we differentiate each term separately.
Step 1: Apply the Product Rule:
[tex]\(f'g + fg'\)[/tex].
Here, [tex]\(f(x) = -3x^5\) and \(g(x) = -7x^5\).[/tex]
Step 2: Find [tex]\(f'(x)\) and \(g'(x)\)[/tex] by differentiating [tex]\(f(x)\) and \(g(x)\)[/tex] with respect to [tex]\(x\)[/tex].
Step 3:
[tex]f'(x) &= \frac{d}{dx}(-3x^5) \\[/tex]
[tex]&= -3 \cdot 5x^{5-1} \\[/tex]
[tex]&= -15x^4[/tex]
[tex]g'(x) &= \frac{d}{dx}(-7x^5) \\[/tex]
[tex]&= -7 \cdot 5x^{5-1} \\[/tex]
[tex]&= -35x^4[/tex]
Step 4: Now, plug [tex]\(f(x)\), \(f'(x)\), \(g(x)\), and \(g'(x)\)[/tex] into the Product Rule formula [tex]\(f'g + fg'\)[/tex]:
[tex]g'(x) &= (-15x^4)(-7x^5) + (-3x^5)(-35x^4) \\[/tex]
[tex]&= 105x^9 + 105x^9 \\[/tex]
[tex]&= 105x^9 - 105x^9 \\[/tex]
[tex]&= -105x^9[/tex]
Step 5: Simplify the result:
[tex]g'(x) &= -105x^9 + 35x^9 \\[/tex]
[tex]&= -105x^5 + 35x^{10}[/tex]
So, [tex]\(g'(x) = -105x^5 + 35x^{10}\)[/tex].
Therefore, the correct option is c) [tex]-105x^5 + 35x^{10}[/tex]
