Evaluate the following integral, $\int\sqrt{4-\sqrt{x}}dx$
$\int \sqrt{4-\sqrt{x}}dx=\int \sqrt{2^2-(x^{1/4})^2}dx$ Considering the common subsitution for $a^2-x^2$, let $x^{1/4}=2\sin t$ $x=16\sin^4t$$\int dx=\int 64\sin^3t\cos t dt$
Therefore by subsitution, we have $\int \sqrt{4-\sqrt{x}}dx=\int \sqrt{{4-(2\sin t)^2}}(64 \sin^3t\cos t)dt=\int \sqrt{4\cos^2 t}(64 \sin^3t\cos t) dt=\int128\cos^2t\sin^3 t dt=\int 128(\cos^2 t)*(1-cos^2t)\sin tdx=128\int \cos^2t\sin t-\cos^4 t\sin t dt=128(-1/3\cos^3 t+1/5\cos^5t)+C$
Is there any mistake in my workings? This is a very important piece of work for me, and I was hoping SE could check it for me. Thanks!
I suspect something is very wrong but I can't dig the mistake...i.e I tested it for a definite integral and it got me a different answer.
P.S Sorry about the messy typing. I am rather new to LaTex.