2

I am trying to understand a general case for the substitution rule.
My real case is this:

$$\int{}(\sqrt{9-t^2})(-2t)dt$$ Making a generality. If I have an integral in the form of:

$$\int{ab}$$

Where a need to be treated by the SUBSTITUTION RULE and b can be compute directly.

Procedure:

a. compute the integral for u

a=u

$\int{(u)}du=\frac{u^2}{2}du$

After that I need to make the substitution back. But what happens whit the b?

Option 1. be remains the same (don't look for me like the correct answer)

$\frac{u^2}2(b)$

or,

Option 2. (I don't feel absolute comfortable with this)

$\frac{u^2}2(\frac{b^2}2)$

or,

Option 3.

None of the above If is the case please explain the correct path

GambitSquared
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Haizum Skallah
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2 Answers2

1

What you seem to be neglecting is accounting for $du$ when you permform substitution.

E.g.

$$\int{}\sqrt{\color{blue}{\bf(9-t^2)}}\underbrace{(-2t)dt}_{du}$$

Here, we can easily choose $\color{blue}{\bf u = 9 - t^2}$. But note, if we do this, we also need to substitute the equivalent of $-2t\,dt$, as well. We are fortunate with this integrand because it so happens that $$\color{blue}{\bf u = 9 - t^2} \implies \color{red}{\bf du = -2t \,dt}$$

This gives us $$\int \sqrt{(\color{blue}{\bf 9 - t^2})} \color{red}{\bf (-2t \,dt)} = \int \color{blue}{\bf u}^{1/2} \color{red}{\bf du} = \dfrac {u^{3/2}}{3/2} + C = \dfrac 23u^{3/2} + C $$ $$ = \frac 23 (9 - t^2)^{3/2} + C$$

Note that your integral is of the form $$\int \sqrt{9 - t^2}(9-t^2)'\,dt = \int g(f(t)) f'(t) \,dt$$ and our corresponding substitutions are $u = f(t), \; du = f'(t)\,dt$, giving us $$\int g(u) \,du$$

amWhy
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0

Suppose you know that $\,F'(x)=f(x)\,$ and that $\,g\,$ is a differentiable function, then

$$\int g'(x)\,f(g(x))\,dx=F(g(x))+C(=\text{a constant})$$

In your example, $\;-2t= \left(9-t^2\right)'\;$ , and since $\;\int \sqrt x\,dx=\frac23 x^{3/2}+C\;$ , we get

$$\int-2t\sqrt{9-t^2}dt=\frac23(9-t^2)^{3/2}+C$$

DonAntonio
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