What's the integral of ∫sec⁡xtan⁡xdx?

∫sec⁡(x)tan⁡(x)dx=sec⁡(x)+C Here is the explanation: ∫sec⁡(x)tan⁡(x)dx=∫1cos⁡(x)×sin⁡(x)cos⁡(x)dx=∫sin⁡(x)(cos⁡(x))2dx=1cos⁡(x)+C=sec⁡(x)+C

u=sec⁡(x)du=(1cos⁡(x))′=−1cos2(x)×(−sin⁡(x)dx)=sin⁡(x)cos2(x)dx=tan⁡(x)sec⁡(x)dxThus,∫tan⁡(x)sec⁡(x)dx=∫du=u=sec⁡(x)+C

∫sec⁡(x)tan⁡(x)=∫d(sec⁡(x))=sec⁡(x)+C

∫sec⁡(x)tan⁡(x)dx=sec⁡(x)+C Here is the explanation: ∫sec⁡(x)tan⁡(x)dx=∫1cos⁡(x)×sin⁡(x)cos⁡(x)dx=∫sin⁡(x)(cos⁡(x))2dx=1cos⁡(x)+C=sec⁡(x)+C

u=sec⁡(x)du=(1cos⁡(x))′=−1cos2(x)×(−sin⁡(x)dx)=sin⁡(x)cos2(x)dx=tan⁡(x)sec⁡(x)dxThus,∫tan⁡(x)sec⁡(x)dx=∫du=u=sec⁡(x)+C

∫sec⁡(x)tan⁡(x)=∫d(sec⁡(x))=sec⁡(x)+C

"What's the integral of ∫sec⁡xtan⁡xdx?" was answered by our community

burkinaval1b

2021-12-17

What's the integral of

\int \sec x \tan x d x

Medicim6

Beginner2021-12-18Added 33 answers

\int \sec (x) \tan (x) d x = \sec (x) + C

Here is the explanation:

\int \sec (x) \tan (x) d x = \int \frac{1}{\cos (x)} \times \frac{\sin (x)}{\cos (x)} d x = \int \frac{\sin (x)}{{(\cos (x))}^{2}} d x = \frac{1}{\cos (x)} + C = \sec (x) + C

Rita Miller

Beginner2021-12-19Added 28 answers

u = \sec (x)

d u = {(\frac{1}{\cos (x)})}^{'} = - \frac{1}{\cos^{2} (x)} \times (- \sin (x) d x) = \frac{\sin (x)}{\cos^{2} (x)} d x = \tan (x) \sec (x) d x

Thus,

\int \tan (x) \sec (x) d x = \int d u = u = \sec (x) + C

nick1337

Expert2021-12-28Added 777 answers

$\int \sec (x) \tan (x) = \int d (\sec (x)) = \sec (x) + C$

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