Evaluating Trig' Integrals

  • Thread starter dx/dy=?
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  • #1
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Main Question or Discussion Point

Can someone please help me in evaluating the following integral?

Im really bad with trig functions, and have been trying to work this out for ages.

[tex]\int \frac{sin x}{cos^2 x} dx[/tex]

Is there a way I can evaluate for [tex]\cos^2 x[/tex] ??????
Or is there something Im missing altogether?



Also,

Can someone please tell me how:
[tex]\int \frac{sec\theta}{cos\theta} dx =\ tan\theta + C[/tex]
????????????????
 
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Answers and Replies

  • #2
TD
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dx/dy=? said:
[tex]\int \frac{sin x}{cos^2 x} dx[/tex]
Substitution: [itex]y = \cos x \Leftrightarrow dy = - \sin xdx[/itex] gives:

[tex]\int {\frac{{\sin x}}
{{\cos ^2 x}}dx} = \int {\frac{{ - 1}}
{{y^2 }}dy = } \frac{1}
{y} + C = \frac{1}
{{\cos x}} + C[/tex]

dx/dy=? said:
Can someone please tell me how:
[tex]\int \frac{sec\theta}{cos\theta} dx =\ tan\theta = C[/tex]
If you know that :
[tex]\sec \theta = \frac{1}{{\cos \theta }}[/tex]

and

[tex]\frac{{d\left( {\tan \theta } \right)}}
{{d\theta }} = \frac{1}
{{\cos ^2 \theta }}[/tex]

then... :smile:
 
  • #3
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Thanks,

For the first integral [tex]\int \frac{sin x}{cos^2 x} dx[/tex] was thinking more along the lines of multiplying out before evaluatng, to try to obtain a more simple integral to evaluate,
but your way is obviously better.

Is there a way this can be evaluated without substitution?
Or is there no function which has [tex]\cos^2 x[/tex] as its derivative?

Thanks again for the help.
 
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  • #4
GCT
Science Advisor
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why would you want cos^2@ as a derivative?
 
  • #5
TD
Homework Helper
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dx/dy=? said:
Is there a way this can be evaluated without substitution?
Or is there no function which has [tex]\cos^2 x[/tex] as its derivative?

Thanks again for the help.
The easiest way is certainly using that substitution.

I don't really see why you'd want to know an anti-derivative for [tex]\cos^2 x[/tex], if you do: just compute it, use:
[tex]\cos \left( {2x} \right) = 2\cos ^2 x - 1 \Leftrightarrow \cos ^2 x = \frac{{1 + \cos \left( {2x} \right)}}
{2}[/tex]

I don't see how that would help though, you don't have [tex]\cos^2 x[/tex] in the integral but [tex]\frac{1}
{{\cos ^2 x}}[/tex], whose anti-derivative is of course [tex]\tan x[/tex]
 
  • #6
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1
[tex]\int \frac{sin x}{cos^2 x} dx = \int \sec x \tan x dx = \sec x + C [/tex]
 
  • #7
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Thanks very much for all your help everyone.

Im slowly getting the hang of Calculus, but im not at all good with derivatives of trig functions.
Ill get there eventually.
 

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