chi tiết tính nguyên hàm `\int`( $\frac{1}{cos^{2}x}$)dx= `\int`( $\frac{1}{cos^{2}(ax+b)}$)dx= `\int`( $\frac{1}{sin^{2}x}$)dx= `\int`( $\frac{1}{sin^{2}(ax+b)}$)dx=

Đáp án+Giải thích các bước giải:

$\circledast \displaystyle \int \dfrac{1}{\cos^2x} \, dx=\tan x+C\\ \circledast =\displaystyle \int \dfrac{1}{\cos^2(ax+b)} \, dx (a \ne 0)\\ =\dfrac{1}{a}\displaystyle \int \dfrac{a}{\cos^2(ax+b)} \, dx\\=\dfrac{1}{a}\displaystyle \int \dfrac{d(ax+b)}{\cos^2(ax+b)} \, \\=\dfrac{1}{a} \tan (ax+b)+C\\\circledast \displaystyle \int \dfrac{1}{\sin^2x} \, dx=-\cot x+C\\\circledast \displaystyle \int \dfrac{1}{\sin^2(ax+b)} \, dx (a \ne 0) \\=\dfrac{1}{a}\displaystyle \int \dfrac{a}{\sin^2(ax+b)} \, dx\\=\dfrac{1}{a}\displaystyle \int \dfrac{d(ax+b)}{\sin^2(ax+b)}\\ =-\dfrac{1}{a}\cot (ax+b)+C.$