int tanx/ (sqrt2 cotx) = কত ? 

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

উত্তর: \(\frac{1}{\sqrt{2}} (\tan x - x) + c\)

ব্যাখ্যা:

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

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

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

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

সুতরাং, \(\frac{1}{\sqrt{2}} \left( \int \sec^2 x dx - \int 1 dx \right) = \frac{1}{\sqrt{2}} (\tan x - x) + c\)

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

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