Integrating $\sqrt{a^2-x^2}$, $\sqrt{a^2+x^2}$, $\sqrt{x^2-a^2}$ and $(px+q)\sqrt{ax^2+bx+c}$

$\int\sqrt{a^2-x^2}\,dx$. Treat as a product with $1$: by parts with $u=\sqrt{a^2-x^2},\ v=1$, $I=x\sqrt{a^2-x^2}-\int\dfrac{-x}{\sqrt{a^2-x^2}}\cdot x\,dx=x\sqrt{a^2-x^2}+\int\dfrac{x^2}{\sqrt{a^2-x^2}}\,dx$. Write $x^2=a^2-(a^2-x^2)$: the remaining integral splits into $a^2\int\dfrac{dx}{\sqrt{a^2-x^2}}-\int\sqrt{a^2-x^2}\,dx=a^2\sin^{-1}\dfrac xa-I$. So $2I=x\sqrt{a^2-x^2}+a^2\sin^{-1}\dfrac xa+c_1$, giving the standard result $I=\dfrac x2\sqrt{a^2-x^2}+\dfrac{a^2}2\sin^{-1}\dfrac xa+c$.

By the identical self-referential technique, $\int\sqrt{a^2+x^2}\,dx=\dfrac x2\sqrt{a^2+x^2}+\dfrac{a^2}2\log\left(x+\sqrt{x^2+a^2}\right)+c$, and $\int\sqrt{x^2-a^2}\,dx=\dfrac x2\sqrt{x^2-a^2}-\dfrac{a^2}2\log\left(x+\sqrt{x^2-a^2}\right)+c$.

$\int x\sin^{-1}x\,dx$. LIATE puts $u=\sin^{-1}x$: $I=\sin^{-1}x\cdot\dfrac{x^2}2-\displaystyle\int\dfrac{1}{\sqrt{1-x^2}}\cdot\dfrac{x^2}2\,dx$. Write $x^2=1-(1-x^2)$: the remaining integral becomes $\int\dfrac{dx}{\sqrt{1-x^2}}-\int\sqrt{1-x^2}\,dx=\sin^{-1}x-\left[\dfrac x2\sqrt{1-x^2}+\dfrac12\sin^{-1}x\right]$. Combining: $I=\dfrac12x^2\sin^{-1}x+\dfrac14x\sqrt{1-x^2}-\dfrac14\sin^{-1}x+c$.

$\int\cos^{-1}\sqrt x\,dx$ (Activity). Put $\sqrt x=t$, $x=t^2$, $dx=2t\,dt$: $I=\int\cos^{-1}t\cdot2t\,dt$, which reduces to the pattern of the previous example ($t\cos^{-1}t$ integrated by parts, $u=\cos^{-1}t$) and can be finished the same way. …