I'm having problems integrating

$\displaystyle

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

$

the answer in the book is

$\displaystyle

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

$

I can't see how this answer is obtained.

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- Jul 5th 2006, 12:12 PMmacca101integration
I'm having problems integrating

$\displaystyle

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

$

the answer in the book is

$\displaystyle

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

$

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

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

When you integrate this, the negatives cancel out and you get $\displaystyle \frac{1}{u}+C$ or $\displaystyle \frac{1}{\cos(x)}+C$ - Jul 5th 2006, 12:24 PMmacca101
Tremendous :D

Thanks Jameson - Jul 5th 2006, 02:34 PMSoroban
Hello, macca101!

Another method . . .

Quote:

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

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

$

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

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