How do you integrate $\int e^{sec 2 x} sec 2 x tan 2 x d x$ $int e^(sec2x)sec2xtan2xdx$ from $[\frac{π}{3}, \frac{π}{2}]$ $[pi/3,pi/2]$ ?

Question

How do you integrate $\int e^{sec 2 x} sec 2 x tan 2 x d x$ $int e^(sec2x)sec2xtan2xdx$ from $[\frac{π}{3}, \frac{π}{2}]$ $[pi/3,pi/2]$ ?

1 Answer

Impact of this question

3000 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-21T05:36:28

$\frac{1}{2} \cdot (e^{- 1} - e^{- 2})$ $1/2 * (e^(-1) - e^(-2))$

Explanation:

$\frac{d}{d x} (sec 2 x) = \frac{d}{d x} (\frac{1}{cos 2 x}) = - \frac{1}{{(cos 2 x)}^{2}} \cdot (- sin 2 x) \cdot 2$ $d/dx(sec 2x) = d/dx (1/(cos 2x)) = - 1/(cos 2x)^2 * (-sin 2x) * 2$
$= sec 2 x \cdot tan 2 x \cdot 2$ $= sec 2x * tan 2x * 2$

Hence

$\frac{d}{d x} e^{sec 2 x} = e^{sec 2 x} \cdot \frac{d}{d x} (sec 2 x) = e^{sec 2 x} \cdot sec 2 x \cdot tan 2 x \cdot 2$ $d/dx e^(sec 2x) = e^(sec 2x) * d/dx(sec 2x) = e^(sec 2x) * sec 2x * tan 2x * 2$

So

$\int_{\frac{π}{3}}^{\frac{π}{2}} e^{sec 2 x} sec 2 x tan 2 x d x = \frac{1}{2} e^{sec 2 x} ∣_{\frac{π}{3}}^{\frac{π}{2}} = \frac{1}{2} \cdot (e^{- 1} - e^{- 2})$ $int_(pi/3)^(pi/2) e^(sec2x)sec2xtan2xdx = 1/2 e^(sec 2x)|_(pi/3)^(pi/2) = 1/2 * (e^(-1) - e^(-2))$

Science

Math

Humanities

... and beyond

How do you integrate $\int e^{sec 2 x} sec 2 x tan 2 x d x$ $int e^(sec2x)sec2xtan2xdx$ from $[\frac{π}{3}, \frac{π}{2}]$ $[pi/3,pi/2]$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate ∫esec2xsec2xtan2xdxint e^(sec2x)sec2xtan2xdx from [π3,π2][pi/3,pi/2]?

1 Answer

Explanation:

Related questions

Impact of this question

How do you integrate $\int e^{sec 2 x} sec 2 x tan 2 x d x$ $int e^(sec2x)sec2xtan2xdx$ from $[\frac{π}{3}, \frac{π}{2}]$ $[pi/3,pi/2]$ ?