Question

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}\end{gathered}$
Answer with proper explanation.


→ Expecting answer from :

★ Moderators
★ Brainly stars and teacher
★ Others best users ​​

Answer 1

$\int \frac{dx}{x} = log(x) + c \\ \int \frac{f \prime(x)}{f(x)} = log \mid \: f(x) \mid + c \\ \int \: tan(x)dx = \int \: \frac{sin(x)}{cos(x)} dx \\ = - log|cos(x)| + c \\ = log |sec(x)| + c \\ the \: same \: rule \: goes \: for \: cot...etc.\\\int {sec}^{2}(x)dx=|tan(x)|+c\\let u=tanx\\ \frac{du}{dx}={sec}^{2}(x)\rightarrow {du}={sec}^{2}(x)dx\\ \int du=|u|+c\\ \therefore tan|x|+c$

Step-by-step explanation:

Am not sure what your question is? But if you are asking about a proof, then you may use Taylor series to prove these integrals...