intsqrt((5-x)/(5+x))dx=?
5sin^-1(x/5)+sqrt(25-x^2)
সমাধান:
ধরি, \(I = \int \sqrt{\frac{5-x}{5+x}} dx\)
আমরা \(\sqrt{\frac{5-x}{5+x}}\) কে \(\sqrt{\frac{(5-x)(5-x)}{(5+x)(5-x)}}\) আকারে লিখতে পারি।
তাহলে, \(I = \int \sqrt{\frac{(5-x)^2}{25-x^2}} dx = \int \frac{5-x}{\sqrt{25-x^2}} dx\)
এখন, \(I = \int \frac{5}{\sqrt{25-x^2}} dx - \int \frac{x}{\sqrt{25-x^2}} dx\)
প্রথম ইন্টিগ্রালটি হলো:
\(I_1 = \int \frac{5}{\sqrt{25-x^2}} dx = 5 \int \frac{1}{\sqrt{5^2-x^2}} dx = 5 \sin^{-1}(\frac{x}{5}) + C_1\)
দ্বিতীয় ইন্টিগ্রালটি হলো:
\(I_2 = \int \frac{x}{\sqrt{25-x^2}} dx\)
ধরি, \(u = 25 - x^2\), তাহলে, \(du = -2x dx\), সুতরাং \(x dx = -\frac{1}{2} du\)
তাহলে, \(I_2 = \int \frac{-\frac{1}{2}}{\sqrt{u}} du = -\frac{1}{2} \int u^{-\frac{1}{2}} du = -\frac{1}{2} \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C_2 = -\sqrt{u} + C_2 = -\sqrt{25-x^2} + C_2\)
সুতরাং, \(I = I_1 - I_2 = 5 \sin^{-1}(\frac{x}{5}) - (-\sqrt{25-x^2}) + C\)
\(I = 5 \sin^{-1}(\frac{x}{5}) + \sqrt{25-x^2} + C\)
অতএব, \(\int \sqrt{\frac{5-x}{5+x}} dx = 5 \sin^{-1}(\frac{x}{5}) + \sqrt{25-x^2} + C\) 🎉
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