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\begin{equation*}\frac{x^3}{(x^2 +1)^{5/2}} = \frac{x^3 + x - x}{(x^2 + 1)^{5/2}} = \frac{x(x^2+1)}{(x^2 +1)^{5/2}} - \frac{x}{(x^2 +1)^{5/2}} = \frac{x}{(x^2+1)^{3/2}} - \frac{x}{(x^2 +1)^{5/2}}\end{equation*}
\displaystyle[br]\begin{align*}I + J &= \int\left(\sqrt{\tan x} + \sqrt{\cot x}\right) \;\mathrm{d}x \\&= \sqrt{2} \int\frac{\sin x + \cos x}{\sqrt{\sin 2x}} \;\mathrm{d}x \\[5pt]&= \sqrt{2} \int\frac{(\sin x - \cos x)'}{\sqrt{1-(\sin x - \cos x)^2}} \;\mathrm{d}x \\[5pt]&= \sqrt{2} \sin^{-1}(\sin x - \cos x) + \mathcal{C}_1 \tag{1} \\\end{align*}
\displaystyle[br]\begin{align*}I - J&= \int\left(\sqrt{\tan x} - \sqrt{\cot x}\right) \;\mathrm{d}x \\&= \sqrt{2} \int\frac{(\sin x - \cos x)}{\sqrt{\sin 2x}} \;\mathrm{d}x \\&= -\sqrt{2} \int\frac{(\sin x + \cos x)'}{\sqrt{(\sin x + \cos x)^2 - 1}} \;\mathrm{d}x \\&= -\sqrt{2} \ln\left|(\sin x + \cos x) + \sqrt{(\sin x + \cos x)^2 - 1}\right| + \mathcal{C}_2 \tag{2} \\\end{align*}
\displaystyle\begin{equation*} I = \frac{1}{\sqrt{2}} \sin^{-1}(\sin x - \cos x) - \frac{1}{\sqrt{2}} \ln\left|\sin x + \cos x + \sqrt{\sin 2x} \vphantom{x^{x^x}} \right| + \mathcal{C}\end{equation*}
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year 13 gyg journal : trying not to become an academic victim 🤡📖Last reply 1 day ago
year 13 gyg journal : trying not to become an academic victim 🤡📖