# integration

• July 5th 2006, 12:12 PM
macca101
integration
I'm having problems integrating

$
\int\frac {sin(x)}{cos^2(x)}dx
$

the answer in the book is

$
\frac{1}{cos(x)}+c
$

I can't see how this answer is obtained.
• July 5th 2006, 12:16 PM
Jameson
Let u=cos(x) and du=-sin(x)

This becomes $\int \frac{1}{u^2}(-1du)$

When you integrate this, the negatives cancel out and you get $\frac{1}{u}+C$ or $\frac{1}{\cos(x)}+C$
• July 5th 2006, 12:24 PM
macca101
Tremendous :D

Thanks Jameson
• July 5th 2006, 02:34 PM
Soroban
Hello, macca101!

Another method . . .

Quote:

$\int\frac{\sin(x)}{\cos^2(x)}\,dx$

The answer in the book is: . $\frac{1}{\cos(x)}+C
$

I'm surprised that they left the answer like that . . .

We have: . $\int\frac{\sin x}{\cos^2x}\,dx\;=\;\int\frac{1}{\cos x}\cdot\frac{\sin x}{\cos x}\,dx\;=$ $\int\sec x\tan x\,dx \;= \;\sec x + C$