e-tan-sec-sin-d-

Question Number 36459 by rahul 19 last updated on 02/Jun/18

\int e^{\tan θ} (\sec θ - \sin θ) d θ = ?

Answered by ajfour last updated on 02/Jun/18

l e t \tan θ = t

\Rightarrow I = \int e^{t} (\sec θ - \sin θ) \cos^{2} θ d t

= \int e^{t} (\cos θ - \frac{\sin θ}{1 + \tan^{2} θ}) d t

= \int e^{t} (\frac{1}{\sqrt{1 + t^{2}}} - \frac{t}{{(1 + t^{2})}^{3 / 2}}) d t

= \int e^{t} [\frac{1}{\sqrt{1 + t^{2}}} + \frac{d}{d t} (\frac{1}{\sqrt{1 + t^{2}}})] d t

= \frac{e^{t}}{\sqrt{1 + t^{2}}} + c

I = e^{\tan θ} \cos θ + c .

Commented by rahul 19 last updated on 02/Jun/18

Thank you sir ��

Answered by MJS last updated on 02/Jun/18

lol, tricky \dots

\int e^{\tan θ} (\sec θ - \sin θ) d θ =

= \int e^{\tan θ} \sec θ d θ - \int e^{\tan θ} \sin θ d θ =

[\begin{matrix} f^{'} = \sin θ \Rightarrow f = - \cos θ \\ g = e^{\tan θ} \Rightarrow g^{'} = e^{\tan θ} \sec^{2} θ \\ \int f^{'} g = f g - \int {f g}^{'} \end{matrix}]

= \int e^{\tan θ} \sec θ d θ + e^{\tan θ} \cos θ - \int e^{\tan θ} \cos θ \sec^{2} θ d θ =

= e^{\tan θ} \cos θ + C

Commented by rahul 19 last updated on 02/Jun/18

How fg = \int f^{^{'}} g + \int {fg}^{'} ? ?

Integration by parts says,

\int fg = f \int g - \int f^{'} \int g .

Commented by MJS last updated on 02/Jun/18

h t t p s : / / e n . w i k i p e d i a . o r g / w i k i / I n t e g r a t i o n_b y_p a r t s

Commented by rahul 19 last updated on 02/Jun/18

Sir if you want to integrate say

\int x \sin x then g = \sin x, f = x . (ILATE)

\Rightarrow \int g = - \cos x, f^{'} = 1 .

\int fg = f \int g - \int f^{'} \int g

\Rightarrow x (- \cos x) - \int - (\cos x)

\Rightarrow - x \cos x + \sin x + c .

Commented by MJS last updated on 02/Jun/18

we come from differentation :

{(u v)}^{'} = u^{'} v + {u v}^{'}

now integrate both sides :

u v = \int u^{'} v + \int {u v}^{'}

{that}^{'} s how I learned it \dots

Commented by rahul 19 last updated on 02/Jun/18

Is this also called integration by parts ?

Anyways, thanks!!