int(dx)/(cos^2xsqrt(tanx))=?  

Explanation:

Another Explanation (5): ```html

সমাধান

ধরি, \(I = \int \frac{dx}{\cos^2 x \sqrt{\tan x}}\) আমরা জানি, \(\frac{d}{dx}(\tan x) = \sec^2 x = \frac{1}{\cos^2 x}\) সুতরাং, \(I = \int \frac{\sec^2 x}{\sqrt{\tan x}} dx\) এখন, ধরি \(u = \tan x\), তাহলে \(\frac{du}{dx} = \sec^2 x\) বা, \(du = \sec^2 x dx\) তাহলে, \(I = \int \frac{du}{\sqrt{u}} = \int u^{-\frac{1}{2}} du\) আমরা জানি, \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\) সুতরাং, \(I = \frac{u^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C = \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C = 2\sqrt{u} + C\) \(u\) এর মান বসিয়ে পাই, \(I = 2\sqrt{\tan x} + C\) অতএব, \(\int \frac{dx}{\cos^2 x \sqrt{\tan x}} = 2\sqrt{\tan x} + C\) 🥳🎉 সুতরাং, উত্তর: \(2\sqrt{\tan x}\) ```