int(tanx)/(sqrt2 cotx)  =কত?

প্রশ্ন: \(\int \frac{\tan x}{\sqrt{2} \cot x} dx = ?\)

সমাধান:

আমরা জানি, \(\cot x = \frac{1}{\tan x}\)

সুতরাং, \(\int \frac{\tan x}{\sqrt{2} \cot x} dx = \int \frac{\tan x}{\sqrt{2} \cdot \frac{1}{\tan x}} dx\)

\(= \int \frac{\tan^2 x}{\sqrt{2}} dx\)

\(= \frac{1}{\sqrt{2}} \int \tan^2 x dx\)

আমরা জানি, \(\sec^2 x - \tan^2 x = 1\), সুতরাং \(\tan^2 x = \sec^2 x - 1\)

\(= \frac{1}{\sqrt{2}} \int (\sec^2 x - 1) dx\)

\(= \frac{1}{\sqrt{2}} \left[ \int \sec^2 x dx - \int 1 dx \right]\)

আমরা জানি, \(\int \sec^2 x dx = \tan x\) এবং \(\int 1 dx = x\)

\(= \frac{1}{\sqrt{2}} [\tan x - x] + C\)

\(= \frac{1}{\sqrt{2}} (\tan x - x) + C\)

অতএব, \(\int \frac{\tan x}{\sqrt{2} \cot x} dx = \frac{1}{\sqrt{2}} (\tan x - x) + C\)

সুতরাং, উত্তর: \(\frac{1}{\sqrt{2}} (\tan x - x) + C\) 🎉