intsec^2xcosec^2xdx=? | Address Academy Question

Another Explanation (5): সমাধান: \[ \int \sec^2x \cdot \csc^2x \, dx = ? \] আমরা জানি, \(\sec^2x = 1 + \tan^2x\) এবং \(\csc^2x = 1 + \cot^2x\) তাহলে, \[ \begin{aligned} \int \sec^2x \cdot \csc^2x \, dx &= \int (1+\tan^2x)(1+\cot^2x) \, dx \\ &= \int (1 + \tan^2x + \cot^2x + \tan^2x \cot^2x) \, dx \\ &= \int (1 + \tan^2x + \cot^2x + 1) \, dx \quad [\because \tan x \cot x = 1] \\ &= \int (2 + \tan^2x + \cot^2x) \, dx \\ &= \int ( \sec^2x + \csc^2x) \, dx \quad [\because \sec^2x = 1+\tan^2x, \csc^2x = 1+\cot^2x] \\ &= \int \sec^2x \, dx + \int \csc^2x \, dx \\ &= \tan x - \cot x + c \end{aligned} \] অপর একটি পদ্ধতি: আমরা জানি, \(\sec^2x = \frac{1}{\cos^2x}\) এবং \(\csc^2x = \frac{1}{\sin^2x}\) \[ \begin{aligned} \int \sec^2x \cdot \csc^2x \, dx &= \int \frac{1}{\cos^2x} \cdot \frac{1}{\sin^2x} \, dx \\ &= \int \frac{1}{\sin^2x \cos^2x} \, dx \\ &= \int \frac{4}{4\sin^2x \cos^2x} \, dx \\ &= 4 \int \frac{1}{(2\sin x \cos x)^2} \, dx \\ &= 4 \int \frac{1}{\sin^2 2x} \, dx \\ &= 4 \int \csc^2 2x \, dx \\ &= 4 \cdot \frac{-\cot 2x}{2} + c \\ &= -2 \cot 2x + c \\ &= -2 \cdot \frac{\cot^2x - 1}{2\cot x} + c \\ &= - \frac{\cot^2x - 1}{\cot x} + c \\ &= - \cot x + \frac{1}{\cot x} + c \\ &= - \cot x + \tan x + c \\ &= \tan x - \cot x + c \end{aligned} \] অতএব, \(\int \sec^2x \cdot \csc^2x \, dx = \tan x - \cot x + c\) 🥳🎉