Is this a valid manoeuvre, or does it need further validation?
$$\dfrac{\pi}{4}=\int_0^1 (1-x^2)^{1/2}dx$$ $$=\int_0^1 e^{\ln{(1-x^2)^{1/2}}}dx$$ $$=\int_0^1 e^{\frac12\ln{(1-x^2)}}dx$$
We can then turn the $\ln$ into a power series, and pass this to $e$ as $e^X$ to further expand into another power series, which we can then integrate to give a power series for $\pi/4$.
You can use substitution as mentioned by @achille, or you can integrate by parts, setting: $$u=\sqrt{1-x^2}, \;\mathrm d\mkern1mu v=\mathrm d\mkern1mu x,\enspace\text{whence}\enspace \mathrm d\mkern1mu u=-\frac{x\,\mathrm d\mkern1mu x}{\sqrt{1-x^2}},\;v=x.$$ The formula yields: \begin{align*}\int_0^1\sqrt{1-x^2}\,\mathrm d\mkern1mu x & =x\sqrt{1-x^2}\,\biggl\rvert_0^1+\int_0^1\frac{x^2\,\mathrm d\mkern1mu x}{\sqrt{1-x^2}}=\int_0^1\frac{(x^2-1)\,\mathrm d\mkern1mu x}{\sqrt{1-x^2}}+\int_0^1\frac{\mathrm d\mkern1mu x}{\sqrt{1-x^2}}\\ &=-\int_0^1\sqrt{1-x^2}\,\mathrm d\mkern1mu x +\arcsin x\,\biggl\rvert_0^1=-\int_0^1\sqrt{1-x^2}\,\mathrm d\mkern1mu x + \frac\pi2, \end{align*} whence the result.