Using integration by parts (fixed it)..
It can actually be done without the substitution. And if you know the standard integral for sqrt(1+x^2), you only need IBP once.
Get an integral of secant^5 u, integration by parts twice to arrive at the answer.
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Prove ∫f(x)dx = ∫f(a-x)dx with limits 0 to π/2, using u=a-x.
Get an integral of secant^5 u, integration by parts twice to arrive at the answer.
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LOL
This is not exactly quick, but I think it works :/ I get the feeling there is some nifty manipulation you were hoping for.\sqrt{x-x^2}})
Very nice!This is not exactly quick, but I think it works :/ I get the feeling there is some nifty manipulation you were hoping for.
\sqrt{x}~\sqrt{1-x}}\\\\$Let $u=\sqrt{x}\\\\du=\frac{dx}{2\sqrt{x}}\\\\\sqrt{1-x}=\sqrt{1-u^2}\\\\\int \frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}=2\int \frac{du}{(1+u)\sqrt{1-u^2}}\\\\$Let $u=sin\theta\\\\du=\cos\theta~d\theta\\\\\int \frac{dx}{(1+\sqrt{x})\sqrt{x-x^2}}=2\int\frac{d\theta}{1+\sin\theta}\\\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2\int \frac{1-\sin\theta}{\cos^2\theta}~d\theta\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2\int(\sec^2\theta-\sec\theta\tan\theta)d\theta\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2(\tan\theta-\sec\theta)+c\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2( \frac{u}{\sqrt{1-u^2}}-\frac{1}{\sqrt{1-u^2}})+c\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2(\frac{ \sqrt{x}-1}{\sqrt{1-x}})+c)
I wasn't sure how to latex the limits when subbing into the primitive so I sort of left that part out (I'm used to solving just indefinite integrals on BOS in the past lol)
Yep nice work.I wasn't sure how to latex the limits when subbing into the primitive so I sort of left that part out (I'm used to solving just indefinite integrals on BOS in the past lol)
^2} \ dx \\=\lim_{t \to \infty}\frac{1}{2}\left [ \tan^{-1}x^2 \right ]\\=\lim_{t \to \infty}\frac{1}{2}\tan^{-1}t^2\\=\frac{\pi}{4})
In MathType, there is a dual superscript/subscript template which works effectively. I wonder if there is a similar command in Latex.
