# Definite integrals

• Jun 4th 2009, 12:18 AM
helloying
Definite integrals
$\frac{d}{dx}\tan x=\sec^2 x$
$\frac{d}{dx}\sec x = \sec x\sin x$

Hence evaluate $\int_0^\pi \frac{1 + \sin x}{\cos^2 x} dx$
• Jun 4th 2009, 12:44 AM
Amer
write it like that it easy now

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

$\int sec^2xdx +\int\frac{sinx}{cos^2x}dx$
• Jun 4th 2009, 12:45 AM
pickslides
$\int_0^\pi \frac{1 + \sin (x)}{\cos^2 (x)} dx$

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

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

$\int_0^\pi sec^2 (x) + tan (x)sec(x)dx$
• Jun 4th 2009, 12:48 AM
pickslides
Quote:

Originally Posted by Amer
write it like that it easy now

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

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

$
\int\frac{sinx}{cos^2x}dx \neq \int{sinx}{cos^2x}dx
$
• Jun 4th 2009, 12:53 AM
pickslides
$\frac{d}{dx}\tan (x)=\sec^2( x) \Rightarrow \tan (x)=\int \sec^2( x)dx$

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