Интеграл табыу — математика анализда дифференцил табыу менән нигеҙ операцияһы.
Уң яҡта дифференциал табабыҙ:
d d x ( 1 a arctg x a + C ) = d d x ( 1 a arctg x a ) = 1 a ⋅ d d ( x a ) ( arctg x a ) ⋅ d d x ( x a ) = 1 a ⋅ 1 1 + x 2 a 2 ⋅ 1 a = 1 a 2 ⋅ a 2 + x 2 a 2 = 1 a 2 + x 2 {\displaystyle {d \over dx}\,\left({1 \over a}\,\operatorname {arctg} \,{\frac {x}{a}}+C\right)={d \over dx}\,\left({1 \over a}\,\operatorname {arctg} \,{\frac {x}{a}}\right)={\frac {1}{a}}\cdot {d \over d\left({x \over a}\right)}\left(\operatorname {arctg} {\frac {x}{a}}\right)\cdot {d \over dx}\left({x \over a}\right)={\frac {1}{a}}\cdot {\frac {1}{1+{\frac {x^{2}}{a^{2}}}}}\cdot {\frac {1}{a}}={\frac {1}{a^{2}\cdot {\frac {a^{2}+x^{2}}{a^{2}}}}}={\frac {1}{a^{2}+x^{2}}}}
∫ tg x d x = ∫ sin x cos x d x = − ∫ d ( cos x ) cos x = − ln | cos x | + C {\displaystyle \int \!\operatorname {tg} \,{x}\,dx=\int {\frac {\sin x}{\cos x}}dx=-\int {\frac {d(\cos x)}{\cos x}}=-\ln |\cos x|+C}
∫ ctg x d x = ∫ cos x sin x d x = ∫ d ( sin x ) sin x = ln | sin x | + C {\displaystyle \int \!\operatorname {ctg} \,{x}\,dx=\int {\frac {\cos x}{\sin x}}dx=\int {\frac {d(\sin x)}{\sin x}}=\ln |\sin x|+C}
Иҫбатлау ∫ sech x d x = arctan ( sh x ) + C {\displaystyle \int \operatorname {sech} \,x\,dx=\arctan(\operatorname {sh} \,x)+C} выполним проверкой:
Иҫбатлау ∫ sech x d x = 2 arctg ( e x ) + C {\displaystyle \int \operatorname {sech} \,x\,dx=2\operatorname {arctg} (e^{x})+C} выполним проверкой:
Иҫбатлау ∫ sech x d x = 2 arctg ( th x 2 ) + C {\displaystyle \int \operatorname {sech} \,x\,dx=2\,\operatorname {arctg} \,\left(\operatorname {th} \,{\frac {x}{2}}\right)+C} тикшерәбеҙ: