Hello everyone

 

One of the example questions of the acceptance exam for Med school, is the following:

(http://www.ond.vlaanderen.be/toelatingsexamen/nl/modelvragen/files/2013/modelvragen-wiskunde-2013.pdf "Vraag 7")

 

Given:

[math]\int{\frac{dx}{x}}=\ln{x}[/math] and [math]\int{\cos{x}dx}=\sin{x}[/math]

 

Which one of the following statements is true?

 

[math]\int{\frac{dx}{\cos{x}}}=\ln{\sin{x}}+C[/math]

 

[math]\int{\frac{dx}{\cos{x}}}=\sin{\ln{x}}+C[/math]

 

[math]\int{\frac{dx}{\cos{x}}}=\ln{x}+\frac{1}{\sin{x}}+C[/math]

 

[math]\int{\frac{dx}{\cos{x}}}[/math] cannot be calculated on base of the given.

 

The last option should be the correct one.

 

But what about this:

 

[math]\int{\left[1\cdot \frac{1}{\cos{x}}dx\right]}=\frac{x}{\cos{x}}-\int{\left[x d\left(\frac{1}{\cos{x}}\right)\right]}[/math]

 

[math]\cdots = \frac{x}{\cos{x}}-\int{\left[x\cdot\sin{x}\right]}=\frac{x}{\cos{x}}-\int{\left[x\cdot d\left(-\cos{x}\right)\right]}[/math]

 

[math]\cdots = \frac{x}{\cos{x}}+x\cdot\cos{x}+\int{\cos{x}dx}=\frac{x}{\cos{x}}+x\cdot\cos{x}+\sin{x}+C[/math]

Edited February 19, 201411 yr by Function

You have the right idea with your 1/1 but with the wrong part of the integral
[math]I = \int {\frac{1}{{\cos x}}dx} [/math]

 

[math] = \int {\frac{{\cos x}}{{{{\cos }^2}x}}dx} [/math]

[math] = \int {\frac{{\cos x}}{{1 - {{\sin }^2}x}}dx} [/math]

If we make the substitution sinx=u

 

[math] I = \int {\frac{{du/dx}}{{1 - {u^2}}}} dx = \int {\frac{1}{{1 - {u^2}}}} du[/math]

This has three different (logarithmic and trigonometric) forms.

 

Does this help?

Edited February 19, 201411 yr by studiot

You have the right idea with your 1/1 but with the wrong part of the integral

[math]I = \int {\frac{1}{{\cos x}}dx} [/math]

 

 

[math] = \int {\frac{{\cos x}}{{{{\cos }^2}x}}dx} [/math]

 

[math] = \int {\frac{{\cos x}}{{1 - {{\sin }^2}x}}dx} [/math]

 

If we make the substitution sinx=u

 

[math] I = \int {\frac{{du/dx}}{{1 - {u^2}}}} dx = \int {\frac{1}{{1 - {u^2}}}} du[/math]

 

This has three different (logarithmic and trigonometric) forms.

 

Does this help?

 

Hmm.. I understand what you're doing, but I still don't see why my reasoning wasn't all correctly:

 

[math]\int{u\cdot v' dx}=\int{u\cdot dv} = uv-\int{v du}[/math]

 

So why can't [math]u=\frac{1}{\cos{x}}[/math] and [math]v=x[/math]

Because you have made d(1/cosx) in your line 1 equal to sinx in your line 2 rather than sinx/cos²x

 

So work your integration by parts out again.

Edited February 19, 201411 yr by studiot

Because you have made d(1/cosx) in your line 1 equal to sinx in your line 2 rather than sinx/cos²x

 

So work your integration by parts out again.

 

Ah yes, how could I've forgotten to divide by cos²x...

 

Continuing your method, the answer would be [math]\arctan{u}+C=\arctan{\sin{x}}+C[/math] or [math]\frac{\ln{\left|u\right|}}{2u}+C=\frac{\ln{\left|\sin{x}\right|}}{2\sin{x}}+C[/math]

 

Correct?

Edited February 19, 201411 yr by Function

[math]I = \int {\frac{{du}}{{1 - {u^2}}}} [/math]

[math] = \frac{1}{2}\ln \frac{{1 + u}}{{1 - u}}[/math]

[math] = \frac{1}{2}\ln \frac{{1 + \sin x}}{{1 - \sin x}}[/math]

[math] = \ln |\tan \left( {\frac{\pi }{4} + \frac{x}{2}} \right)|[/math]

[math] = \ln |\sec x + \tan x|[/math]

[math]I = \int {\frac{{du}}{{1 - {u^2}}}} [/math]

[math] = \frac{1}{2}\ln \frac{{1 + u}}{{1 - u}}[/math]

 

[math] = \frac{1}{2}\ln \frac{{1 + \sin x}}{{1 - \sin x}}[/math]

 

[math] = \ln |\tan \left( {\frac{\pi }{4} + \frac{x}{2}} \right)|[/math]

 

[math] = \ln |\sec x + \tan x|[/math]

 

 

 

(+C, of course.) Thank you.