I'm having trouble matching these integrals to their corresponding integral in the table of integrals.
I have this integral:
$$\int_0^2 x^2*\sqrt{4-x^2}$$
Which one of these matches?
I have this integral:
$$u*\sqrt{a+bu}du = \frac{2}{15b^2} * (ebu-2a)(a+bu)^{\frac{3}{2}} + C$$
but wolfram gives me that the answer is $\pi$. Am I using the wrong match? It looks like it since number 54 (in the table below) has a + sign whereas this has a minus sign. Perhaps number 113 is a better match? But this form doesn't 100% correspond…
If I use the integral form:
$$\int u \sqrt{a+bu}$$
Can I use $u = x^2$, $a = 4$, and $b = -1$?
I also have this integral:
$$\int \frac{\sqrt{2y^2-3}}{y^2}$$
I am totally lost on this one. Nothing even looks close?
Here are the tables that are in question:
Best Answer
The reason the table does not work is because it is in terms of $du$, but you are trying to use the substitution $u=x^2$ without converting the $dx$ into $du$ first. If we try to convert this into $du$ using $dx=\frac{1}{2\sqrt u}dx$, we find that:
$$\int_0^2 x^2\sqrt{4-x^2}dx=\int_0^4 \frac{u\sqrt{4-u}}{2\sqrt u}du=\frac{1}{2}\int_0^4 \sqrt{\frac{4-u}{u}}du$$
Now, this doesn't match anything on your table, so I think we should try a different u-substitution. Since we have $\sqrt{4-x^2}$, let's try $x=2\sin u$ since that will get rid of the square root. Remember to substitute $dx=2\cos u$:
$$\int_0^2 x^2\sqrt{4-x^2}dx=\int_0^{\pi/2} 4\sin^2 u\cdot 2\cos u\cdot 2\cos udu=\int_0^{\pi/2} 16(\sin u\cos u)^2du$$
Remember that $\sin u\cos u=\frac{\sin 2u}{2}$:
$$\int_0^{\pi/2} 16(\sin u\cos u)^2du=\int_0^{\pi/2} 4\sin^2(2u)du$$
Let $v=2u$:
$$\int_0^{\pi/2} 4\sin^2(2u)du=\int_0^\pi 2\sin^2vdv$$
Remember that $2\sin^2v=1-\cos(2v)$:
$$\int_0^\pi (1-\cos(2v))dv=v-\frac{\sin(2v)}{2}^\pi_0=(\pi-0)-(0-0)=\pi$$