Definite integrals

• Jun 3rd 2009, 11:18 PM
helloying
Definite integrals
$\displaystyle \frac{d}{dx}\tan x=\sec^2 x$
$\displaystyle \frac{d}{dx}\sec x = \sec x\sin x$

Hence evaluate $\displaystyle \int_0^\pi \frac{1 + \sin x}{\cos^2 x} dx$
• Jun 3rd 2009, 11:44 PM
Amer
write it like that it easy now

$\displaystyle \int\frac{1}{cos^2x}dx+\int\frac{sinx}{cos^2x}dx$

$\displaystyle \int sec^2xdx +\int\frac{sinx}{cos^2x}dx$
• Jun 3rd 2009, 11:45 PM
pickslides
$\displaystyle \int_0^\pi \frac{1 + \sin (x)}{\cos^2 (x)} dx$

$\displaystyle \int_0^\pi \frac{1}{\cos^2 (x)} +\frac{ \sin (x)}{\cos^2 (x)}dx$

$\displaystyle \int_0^\pi sec^2 (x) +\frac{ \tan (x)}{\cos(x)}dx$

$\displaystyle \int_0^\pi sec^2 (x) + tan (x)sec(x)dx$
• Jun 3rd 2009, 11:48 PM
pickslides
Quote:

Originally Posted by Amer
write it like that it easy now

$\displaystyle \int\frac{1}{cos^2x}dx+\int\frac{sinx}{cos^2x}dx$

$\displaystyle \int sec^2xdx +\int{sinx}{cos^2x}dx$

$\displaystyle \int\frac{sinx}{cos^2x}dx \neq \int{sinx}{cos^2x}dx$
• Jun 3rd 2009, 11:53 PM
pickslides
$\displaystyle \frac{d}{dx}\tan (x)=\sec^2( x) \Rightarrow \tan (x)=\int \sec^2( x)dx$

$\displaystyle \frac{d}{dx}\sec( x) = \sec( x)\sin (x) \Rightarrow \sec( x) =\int \sec( x)\sin (x)dx$