Please explain this Q to me :) (1 Viewer)

LoveHateSchool · Oct 20, 2012

So I have to evaluate integral of secx with limits pi/3 and pi/6.

Already shown these derivatives.
secx=secxtanx
secx + tanx=secx (secx + tan x)

Where to from here?

SpiralFlex · Oct 20, 2012

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec x \; dx=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec x * \frac{\sec x +\tan x}{\sec x +\tan x}\;dx$

$=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sec x(\sec x +\tan x)}{\sec x + \tan x}\; dx$

Notice that the derivative of the bottom is the top. So it's a natural logarithm! That is,

$\int \frac{f'(x)}{f(x)}\;dx=\ln f(x)+C$

$\therefore \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec x \; dx=[\ln (\sec x+\tan x)]_{\frac{\pi}{6}}^{\frac{\pi}{3}}$

$=\ln(\sec(\frac{\pi}{3})+\tan(\frac{\pi}{3}))-\ln(\sec(\frac{\pi}{6})+\tan(\frac{\pi}{6}))$

$=\ln(2+\sqrt{3})-\ln (\frac{2\sqrt{3}+3}{3})$

$=\ln (\frac{3(2+\sqrt{3})}{2\sqrt{3}+3})$

Rationalise at your peril

LoveHateSchool · Oct 20, 2012

SpiralFlex said:
$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec x \; dx=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec x * \frac{\sec x +\tan x}{\sec x +\tan x}\;dx$

$=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sec x(\sec x +\tan x)}{\sec x + \tan x}\; dx$

Notice that the derivative of the bottom is the top. So it's a natural logarithm! That is,

$\int \frac{f'(x)}{f(x)}\;dx=\ln f(x)+C$

$\therefore \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sec x \; dx=[\ln (\sec x+\tan x)]_{\frac{\pi}{6}}^{\frac{\pi}{3}}$

$=\ln(\sec(\frac{\pi}{3})+\tan(\frac{\pi}{3}))-\ln(\sec(\frac{\pi}{6})+\tan(\frac{\pi}{6}))$

$=\ln(2+\sqrt{3})-\ln (\frac{2\sqrt{3}+3}{3})$

$=\ln (\frac{3(2+\sqrt{3})}{2\sqrt{3}+3})$

Rationalise at your peril

Thank you so much robot, can you help to explain to me in first line why I times secx by secx + tanx/secx + tanx, is it just to enable integration, as it doesn't change it because you are really multiplying it be one?
But I can follow the rest from there

SpiralFlex · Oct 20, 2012

A great Mathematician once told me "By doing nothing, you're doing something"

Ie.

$\frac{1}{\sqrt{3}}*1=\frac{1}{\sqrt{3}}$

We effectively multiplied it by one to change nothing, but if we write 1 a different way, say

$1=\frac{\sqrt{3}}{\sqrt{3}}$

We can change it's form,

$\frac{1}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3}$

But this is the same number, so we effectively done nothing but we have change its looks.

I times all sides by sec x + tan x/sec x + tan x to enable integration yes, since we need to use the result from the previous question. Hence by doing this, I have on the numerator,

sec x(sec + tan x), which is the result you have provided.

LoveHateSchool · Oct 20, 2012

Thank you so much Spi, the question is good once I realise how you can change it's form, and now I get it

Thank you so much, it totally clicks now!

SpiralFlex · Oct 20, 2012

Live long and prosper, gl on Monday.

Please explain this Q to me :) (1 Viewer)

LoveHateSchool

Retired Sept '14

SpiralFlex

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LoveHateSchool

Retired Sept '14

SpiralFlex

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LoveHateSchool

Retired Sept '14

SpiralFlex

Well-Known Member

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