tìm nguyên hàm `\int`(5+cot²x)dx `\int`(tan²x + cot²x)dx chi tiết

Đáp án+Giải thích các bước giải:

$\displaystyle \int (5+\cot^2 x) \, dx\\ =\displaystyle \int (4+(1+\cot^2 x)) \, dx\\ =4x-\cot x+C\\ \displaystyle \int (\tan^2x+\cot^2 x) \, dx\\ =\displaystyle \int (\tan^2x+1+\cot^2 x+1-2) \, dx\\ =\displaystyle \int (\tan^2x+1) \, dx+\displaystyle \int (\cot^2 x+1) \, dx-2\displaystyle \int \, dx\\ =\tan x-\cot x -2x +C$

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$\cot^2x+1=\dfrac{\cos^2x}{\sin^2x}+\dfrac{\sin^2x}{\sin^2x}=\dfrac{\cos^2x+\sin^2x}{\sin^2x}=\dfrac{1}{\sin^2x}\\ \tan^2x+1=\dfrac{\sin^2x}{\cos^2x}+\dfrac{\cos^2x}{\cos^2x}=\dfrac{\sin^2x+\cos^2x}{\cos^2x}=\dfrac{1}{\cos^2x}.$