integration

[aerogurl2]

1. The problem statement, all variables and given/known data
Hi, i don't really know how I should go about this question. please help me.

Question:
Integrate the following equation:
I = (integration sign from 0 to 2 sqrt3) (x^3)/(16-x^2)^(1/2)

2. Relevant equations



3. The attempt at a solution
I tried to assign u and dv initially like this:

u = (16-x^2)^(1/2)
du = -(16-x^2)^(-1/2) dx

dv = x^3 dx
v = (1/4)(x^4)

so uv - (INT)vdu
= (1/4)(16-x^2)(x^4) + (1/4)(INT)(x^4)/(16-x^4)^(1/2) dx

but it gets more complicated because of x^4, if x^3 in the integration then i could have assigned it as I and then solve it easily, but i think i made a huge mistake somewhere to get it to be like this.

[Curious3141]

With experience, you'll realise that a trigonometric substitution in the denominator works best in cases like these. A big clue is the sqrt(3) in the bounds, this is immediately reminiscent of special angle ratios like sin(pi/3), right?

Anyway, your first objective here is to get rid of the denominator by applying a trig sub. Try x = 4\sin{\theta}. Simplify.

You'll be left with a cubed trig expression to integrate. Do that by parts (\sin^3{\theta} = \sin^2{\theta}\sin{\theta}). It's not difficult, in fact, it can neatly be done in two lines with a little application of some trig identities and a little algebra.

Finally, don't sub x back into the final indefinite integral. Leave everything in terms of theta. Work out the values of theta that correspond to the bounds of the definite integral and evaluate. Much neater.

[ChaoticLlama]

When you see a squareroot, trig substitution can look very tempting, but another method can shorten the work greatly.

For example...


\begin{array}{l}
\int {\frac{{x^3 }}{{\sqrt {16 - x^2 } }}} dx \\
{\rm{let}}{\kern 1pt} {\kern 1pt} {\kern 1pt} u = \sqrt {16 - x^2 } {\kern 1pt} {\kern 1pt} {\kern 1pt} \to x^2 = 16 - u^2 \\
du = \frac{{ - x}}{{\sqrt {16 - x^2 } }}dx \\
\int {\frac{{x^3 }}{{\sqrt {16 - x^2 } }}} dx = \int {\left( {u^2 - 16 } \right)} du \\
\end{array}

[Curious3141]

When you see a squareroot, trig substitution can look very tempting, but another method can shorten the work greatly.

For example...


\begin{array}{l}
\int {\frac{{x^3 }}{{\sqrt {16 - x^2 } }}} dx \\
{\rm{let}}{\kern 1pt} {\kern 1pt} {\kern 1pt} u = \sqrt {16 - x^2 } {\kern 1pt} {\kern 1pt} {\kern 1pt} \to x^2 = 16 - u^2 \\
du = \frac{{ - x}}{{\sqrt {16 - x^2 } }}dx \\
\int {\frac{{x^3 }}{{\sqrt {16 - x^2 } }}} dx = \int {\left( {u^2 - 16 } \right)} du \\
\end{array}



Yes, that's a much neater method! :smile: