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When evaluating $ \int \cos^3(x)dx$, I've only seen that the first step is to recognize that $\cos^3(x)=\cos(x)(\cos^2(x))\Rightarrow \cos(x)(1-\sin^2(x))$, followed by distributing the $\cos(x)$ and then integrating as usual, giving $\sin(x)-\frac{\sin^3(x)}{3}+C$ as the indefinite integral.

My question is if we can use the derivative of $\cos^4(x)$ to evaluate the integral like this: $$\frac{d}{dx}[\cos^4(x)]=-4\cos^3(x)\sin(x)$$

With this, we set up $$\int\cos^3(x)dx=\frac{-4\sin(x)}{-4\sin(x)}\int\cos^3(x)dx$$

And then multiply the $-4\sin(x)$ into the integrand to use the anti-derivative: $$=\frac{1}{-4\sin(x)}\int-4\cos^3(x)\sin(x)dx$$ $$=\frac{1}{-4\sin(x)}(\cos^4(x))+C$$

Which will ultimately give us the indefinite integral as $$\int\cos^3(x)dx=\frac{-\cos^4(x)}{4\sin(x)}+C$$

Is this a legitimate approach? Thank you for your time!

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    $\begingroup$ You cannot pull (or put) a function out of (or in to) an integral. $\endgroup$ May 21 '20 at 17:25
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    $\begingroup$ Another way is to note that $\cos^3(x)=\frac14\cos(3x)+\frac34\cos(x)$ $\endgroup$
    – robjohn
    May 21 '20 at 17:31
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$$\frac{d}{dx}\left(\cos^4(x)\right)=-4\cos^3(x)\sin(x)$$

implies that

$$\int -4\cos^3(x)\sin(x)\,dx = \cos^4(x)+C.$$ While it is true that you can write $$\int\cos^3(x)\,dx=\frac{-4\sin(x)}{-4\sin(x)}\int\cos^3(x)\,dx,$$

it is incorrect to move the $\sin(x)$ into the integrand. The value of $\sin(x)$ depends on the value of $x$, so you cannot move it inside and outside the integral because it is a function which depends on the value of $x$ and therefore isn't a constant.

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I noticed you used a punch of times some property that doesn't exist in integrals: $$\int f(x)g(x)dx = g(x) \int f(x)dx$$ Just try to put $f(x)=g(x)=x$ in the above to see how wrong this is.

The right way you'd want to do something like this is: $$\int \cos^3(x) dx=\int \frac{-4\sin(x)\cos^3(x)}{-4\sin(x)}dx$$ So you either integrate by parts: $$\int \frac{-4\sin(x)\cos^3(x)}{-4\sin(x)}dx = -\frac{\cos^4(x)}{4\sin(x)}-\int \cos^4(x) \cdot \frac{d}{dx}\left(-\frac{1}{4\sin(x)}\right) dx$$ Or by substituting $du=-4\sin(x)\cos^3(x) dx \implies u=\cos^4(x)$, and either way is not easy.

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Recognize $d \sin x $ and it is straight forward:

$$ \int \cos^3 x dx= \int \cos^2 x \cos x\; dx= \int (1-\sin^2 x ) \;d \sin x $$ $$ =\sin x - \dfrac{\sin^3 x}{3} +c $$

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