int1/(cos^2xsqrt(tan x))dx = ? 

ধরি, \(I = \int \frac{1}{\cos^2x \sqrt{\tan x}} dx\)

আমরা জানি, \(\sec^2x = \frac{1}{\cos^2x}\)

সুতরাং, \(I = \int \sec^2x \frac{1}{\sqrt{\tan x}} dx\)

ধরি, \(u = \tan x\)
তাহলে, \(\frac{du}{dx} = \sec^2x\)
সুতরাং, \(du = \sec^2x dx\)

এখন, \(I = \int \frac{1}{\sqrt{u}} du\)
\(I = \int u^{-\frac{1}{2}} du\)

আমরা জানি, \(\int x^n dx = \frac{x^{n+1}}{n+1} + c\)
সুতরাং, \(I = \frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} + c\)
\(I = \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + c\)
\(I = 2\sqrt{u} + c\)

যেহেতু, \(u = \tan x\)
সুতরাং, \(I = 2\sqrt{\tan x} + c\)

অতএব, \(\int \frac{1}{\cos^2x \sqrt{\tan x}} dx = 2\sqrt{\tan x} + c\) 🥳