Understanding Integral Calculus: Solving ∫cos^3xdx with Step-by-Step Guidance

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Homework Help Overview

The discussion revolves around the integral ∫cos^3xdx, situated within the context of integral calculus. Participants are exploring various methods to approach this integral, including trigonometric identities and substitution techniques.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss splitting cos^3x into cos^2x(cosx) and using trigonometric identities, such as cos^2x = 1 - sin^2x. There are mentions of using substitution and references to the triple angle formula for cosine. Some participants express uncertainty about the necessity of formal substitution.

Discussion Status

The discussion is active, with multiple approaches being explored. Some participants have provided guidance on using trigonometric identities and substitution, while others question the need for formal methods. There is no explicit consensus on a single approach, but various lines of reasoning are being considered.

Contextual Notes

Participants are self-studying integral calculus, which may influence their familiarity with the concepts and techniques being discussed. There is a mention of homework constraints, but specific rules or limitations are not detailed.

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Homework Statement


∫cos^3xdx


Homework Equations




The Attempt at a Solution


I started teaching myself integral calculus yesterday and I had no idea what to do for this problem... I tried splitting cos^3x into cos^2x(cosx) but that didn't work. How do you solve this?
 
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You need to split it the way you did it and use trigonometry. [itex]\cos^2 x = 1-\sin^2 x[/itex] plus substitution.
 
guitarphysics said:

Homework Statement


∫cos^3xdx


Homework Equations




The Attempt at a Solution


I started teaching myself integral calculus yesterday and I had no idea what to do for this problem... I tried splitting cos^3x into cos^2x(cosx) but that didn't work. How do you solve this?

[itex]\displaystyle \int cos^n {x} \ dx = \frac{sin{x} \ cos^{n-1}{x}}{n} + \frac{n-1}{n} \int cos^{n-2}x \ dx[/itex].

Alternatively, [itex]\displaystyle \int cos^3 {x} \ dx = \int cos^2 x \ cosx \ dx = \int (1-sin^2 x) cos x \ dx[/itex]. From here, you could use substitution.
 
Do you know about the triple angle formulas? Maybe you've come across it in trigonometry:
$$\cos 3\theta =4\cos^3 \theta - 3\cos \theta$$
Re-arranging:
$$cos^3 \theta = \frac{1}{4} (\cos 3\theta + 3\cos \theta)$$
Therefore,
$$\int cos^3 \theta\,.d\theta= \frac{1}{4} \int (\cos 3\theta + 3\cos \theta)\,.d\theta$$
 
Mandelbroth said:
[itex]\displaystyle \int cos^3 {x} \ dx = \int cos^2 x \ cosx \ dx = \int (1-sin^2 x) cos x \ dx[/itex]. From here, you could use substitution.
... or just use cos x dx = d sin x.
 
Yes, that's what madelbroth meant.
 
HallsofIvy said:
Yes, that's what madelbroth meant.
I know, just saying you don't have to go through a formal substitution, adjusting limits.
 

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