Question

Evaluate the Integral.

Answer 1

\[\int\limits_{}^{}\frac{ x^2 }{ \sqrt{49-x^2} }\]

Answer 2

trig sub

Answer 3

In particular, try substituting \(x=7\sin t\) (or \(\cos t\)), so \(\mathrm dx=7\cos t\,\mathrm dt\) and
\[\int\frac{x^2}{\sqrt{49-x^2}}\,\mathrm dx=\int\frac{(7\sin t)^2}{\sqrt{49-(7\sin t)^2}}(7\cos t)\,\mathrm dt=49\int\sin^2t\,\mathrm dt\]

Answer 4

@HolsterEmission Would it end up being \[49\int\limits_{}^{}1-\cos^2t\]
I know it would end up being \[49t\] but how would i solve for cos squared?

Answer 5

That's more work then necessary, but it also works.

Let's just consider the squared sine term. You might be familiar with what's usually referred to as the half-angle identity:
\[\sin^2t=\frac{1-\cos2t}{2}\]This is all you need to simplify the integrand.

Alternatively, if you do elect to go through with writing \(\sin^2t=1-\cos^2t\), there's a similar identity you can use:
\[\cos^2t=\frac{1+\cos2t}{2}\]So
\[\int\sin^2t\,\mathrm dt=\frac{1}{2}\int(1-\cos2t)\,\mathrm dt\]or
\[\begin{align*}\int(1-\cos^2t)\,\mathrm dt&=\int\left(1-\frac{1+\cos2t}{2}\right)\,\mathrm dt\\[1ex]&=\int\frac{2-1-\cos2t}{2}\,\mathrm dt\\[1ex]&=\frac{1}{2}\int(1-\cos2t)\,\mathrm dt\end{align*}\]which goes to show that either approach gives the same solution.

Answer 6

So then it would be \[\frac{ 49 }{ 2 }t-\frac{ 49 }{ 4}\sin2t+C\]
But then i have to solve for t, so \[t= \sin^{-1} (\frac{ x }{ 7 })\]
\[\frac{ 49 }{ 2 } \sin^{-1} (\frac{ x }{ 7 })-\frac{ 49 }{ 4 }\sin(2(\sin^{-1} ( \frac{ x }{ 7 }))+C\]

Answer 7

Yup, good so far.

For the next step, the double angle identity for sine will help:
\[\sin2t=2\sin t\cos t\]so
\[\sin\left(2\sin^{-1}\frac{x}{7}\right)=2\sin\left(\sin^{-1}\frac{x}{7}\right)\cos\left(\sin^{-1}\frac{x}{7}\right)\]

Answer 8

Nice work above. Here how wolframalpha did it. Very similar to the work before

Answer 9

Would the sin and the arc sin cancel each other out?

Answer 10

Under certain conditions, and for your purposes, yes.

Answer 11

Not to say that cancellation would yield \(1\), but rather in the same sense that \(f(f^{-1}(x))=x\). So \(\sin\left(\sin^{-1}\dfrac{x}{7}\right)=\dfrac{x}{7}\).

Answer 12

What would happen with the \[\cos(\sin^{-1} (\frac{ x }{ 7 }))\]

Answer 13

For that, you need to use some trig properties. In a reference triangle, \(\sin^{-1}\dfrac{x}{7}\) is some angle whose sine is \(\dfrac{x}{7}\), so you have something like this:

How do you find the missing side? You need it to know what the cosine of this angle is.