# Integration with trig functions

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• Aug 11th 2010, 01:07 PM
bobsanchez
Integration with trig functions
Evaluate the integral sinx/cos^2x dx from 0 to pi/3.

So I've gotten it to tanxcosx, but I don't know the indefinite integral of these, and I couldn't find it in my book. Can anyone help?
• Aug 11th 2010, 01:13 PM
yeKciM
Quote:

Originally Posted by bobsanchez
Evaluate the integral sinx/cos^2x dx from 0 to pi/3.

So I've gotten it to tanxcosx, but I don't know the indefinite integral of these, and I couldn't find it in my book. Can anyone help?

you should get under integral $\sec{x} \tan{x} \; dx$ .....
• Aug 11th 2010, 01:14 PM
Plato
$\displaystyle \int {\frac{{\sin (x)dx}}
{{\cos ^2 (x)}} = \frac{{ 1}}
{{\cos (x)}}}$
• Aug 11th 2010, 01:26 PM
bobsanchez
Ah okay, let me work the problem. Thanks.
• Aug 11th 2010, 01:28 PM
bobsanchez
-1?
• Aug 11th 2010, 01:31 PM
yeKciM
i got result 1 :D
• Aug 11th 2010, 01:39 PM
bobsanchez
-1/cos(60) = -2

-1/cos(0) = -1

-2 - (-1) = -2 + 1 = -1

Did I mess up somewhere in there or enter it wrong?
• Aug 11th 2010, 01:44 PM
TheCoffeeMachine
Let $u = \cos{x}$ to get Plato's result (without the typo minus).
• Aug 11th 2010, 01:46 PM
yeKciM
Quote:

Originally Posted by bobsanchez
-1/cos(60) = -2

-1/cos(0) = -1

-2 - (-1) = -2 + 1 = -1

Did I mess up somewhere in there or enter it wrong?

$\displaystyle \int _0 ^{\frac {\pi}{3} } \frac {\sin{x}}{\cos^2{x}} \; dx = \int _0 ^{\frac {\pi}{3} } \sec{x} \tan{x} \; dx =$

$\displaystyle u = \sec {x}$ and now the $\displaystyle du= \tan{x}\sec{x}\; dx$

after that ... i just got

$\displaystyle \int _0 ^{\frac {\pi}{3} }\;du = u \big{|} _0 ^{\frac {\pi}{3} }= \sec{x} \big{|} _0 ^{\frac {\pi}{3} }$

and with limits you get "1" :D
• Aug 11th 2010, 06:33 PM
bobsanchez
Evaluate the integral sec2xtan2x dx.
sec2x + C

This is another good one. My first try resulted in sec2x + C, but was wrong.
• Aug 11th 2010, 06:40 PM
Prove It
Quote:

Originally Posted by bobsanchez
Evaluate the integral sinx/cos^2x dx from 0 to pi/3.

So I've gotten it to tanxcosx, but I don't know the indefinite integral of these, and I couldn't find it in my book. Can anyone help?

Just use a $u$ substitution...

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

Let $u=\cos{x}$ so that $du = -\sin{x}\,dx$ and the integral becomes

$-\int{u^{-2}\,du}$

$= -\left(\frac{u^{-1}}{-1}\right) + C$

$= \frac{1}{u} + C$

$= \frac{1}{\cos{x}} + C$.

Therefore $\int_0^{\frac{\pi}{3}}{\frac{\sin{x}}{\cos^2{x}}\, dx} = \left[\frac{1}{\cos{x}}\right]_0^{\frac{\pi}{3}}$

$= \left[\frac{1}{\cos{\left(\frac{\pi}{3}\right)}}\right] - \left[\frac{1}{\cos{(0)}}\right]$

$= \frac{1}{\frac{1}{2}} - \frac{1}{1}$

$= 2 - 1$

$= 1$.
• Aug 11th 2010, 10:19 PM
bobsanchez
Quote:

Originally Posted by bobsanchez
Evaluate the integral sec2xtan2x dx.
sec2x + C

This is another good one. My first try resulted in sec2x + C, but was wrong.

Can someone take a quick look at this one please?
• Aug 11th 2010, 10:25 PM
eumyang
Quote:

Originally Posted by bobsanchez
Can someone take a quick look at this one please?

You forgot to use the chain rule when finding du.

\begin{aligned}
u &= \sec 2x \\
du &= 2\sec 2x\, \tan 2x \, dx \\
\frac{1}{2}du &= \sec 2x\, \tan 2x \, dx
\end{aligned}

$\displaystyle \int \sec 2x\,\tan 2x\, dx$
$= \displaystyle \int \frac{1}{2} du$
Can you take it from here?
• Aug 11th 2010, 10:35 PM
bobsanchez
...wow...forgot the chain rule? Really? *sigh*

Anyways, thanks for the help. Alright, so...I just integrate and multiply it by du, correct? So (1/2)sec2x + C?
• Aug 11th 2010, 10:38 PM
eumyang
I could read that as $\sec \frac{2x}{2} + C$, but I know that's not what you meant.

$\displaystyle \int \frac{1}{2} du = \frac{1}{2}u + C = \frac{1}{2}\sec 2x + C$
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