- #1

calcnd

- 20

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Hi, I'm new to this forum. I figure it would be appropriate to briefly introduce myself before asking my question.

My name is Nick. I've graduated high school and I am presently taking the year off before attending university next fall. Having a lot of free time has allowed me to begin learning Calculus II topics that I'll inevitably encounter in the future.

Anyways, on to my question(s):

[tex]\int {{-1}\over{cos^2(x)}} dx[/tex]

[tex]= -\int {{sin^2(x) + cos^2(x)}\over{cos^2(x)}} dx [/tex]

[tex]= -\int {{sin^2(x)}\over{cos^2(x)}} dx -\int {{cos^2(x)}\over{cos^2(x)}}dx [/tex]

[tex]= -\int tan^2(x) dx - \int dx [/tex]

[tex]= ?[/tex]

Alas, I do not know how to properly integrate [tex]tan^2(x)[/tex]

I am relatively confident, however, that my above reasoning is corect. This is because the answer to the following integral is supposed to be:

[tex]\int tan^2(x)dx = tan(x) - x + C[/tex] Meaning:

[tex]\int {{-1}\over{cos^2(x)}} dx[/tex]

[tex]= -\int tan^2(x) dx - \int dx [/tex]

[tex]= -tan(x) + x - C - x + D [/tex]

[tex]= -tan(x) + K [/tex]

Which should be the correct answer. Thus, my problem is in integrating the tangent x squared function.Thank-you in advance,

Nick

Edit: It appears this forum uses a different tex protocol than what I am used to on another forum I frequent. I'm attempting to figure this out, but I do appologize for the confusing mess above.

Edit 2: I believe I found the culprit. The textit code does not apply on this forum. :)

My name is Nick. I've graduated high school and I am presently taking the year off before attending university next fall. Having a lot of free time has allowed me to begin learning Calculus II topics that I'll inevitably encounter in the future.

Anyways, on to my question(s):

[tex]\int {{-1}\over{cos^2(x)}} dx[/tex]

[tex]= -\int {{sin^2(x) + cos^2(x)}\over{cos^2(x)}} dx [/tex]

[tex]= -\int {{sin^2(x)}\over{cos^2(x)}} dx -\int {{cos^2(x)}\over{cos^2(x)}}dx [/tex]

[tex]= -\int tan^2(x) dx - \int dx [/tex]

[tex]= ?[/tex]

Alas, I do not know how to properly integrate [tex]tan^2(x)[/tex]

I am relatively confident, however, that my above reasoning is corect. This is because the answer to the following integral is supposed to be:

[tex]\int tan^2(x)dx = tan(x) - x + C[/tex] Meaning:

[tex]\int {{-1}\over{cos^2(x)}} dx[/tex]

[tex]= -\int tan^2(x) dx - \int dx [/tex]

[tex]= -tan(x) + x - C - x + D [/tex]

[tex]= -tan(x) + K [/tex]

Which should be the correct answer. Thus, my problem is in integrating the tangent x squared function.Thank-you in advance,

Nick

Edit: It appears this forum uses a different tex protocol than what I am used to on another forum I frequent. I'm attempting to figure this out, but I do appologize for the confusing mess above.

Edit 2: I believe I found the culprit. The textit code does not apply on this forum. :)

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