Calculating $\int x dx$ using trigonometric functions

Question

Well, I tried to solve the integral: $$\int x dx$$ using trigonometric functions instead of using the general formula for it. (If $n \neq -1$,$\int x^n dx=\frac{x^{n+1}}{n+1}+C$)

So I gave it shot in this way:

$$\int x dx = \int \sin\theta \cos\theta d\theta = \dfrac{1}{2}\int \sin2\theta d\theta=\dfrac{1}{2}\big(\dfrac{-1}{2}\cos^2\theta\big)+C=\dfrac{-1}{4}(\cos^2\theta-\sin^2\theta)+C$$

$$x=\sin\theta ,\cos\theta=\sqrt {1-x^2}, dx = \cos\theta d\theta$$

Thus we substitute them:

$$\int x dx = \dfrac{1}{2}\int \sin2\theta d\theta=\dfrac{-1}{4}(\cos^2\theta-\sin^2\theta)+C=\dfrac{-1}{4}+\dfrac{x^2}{2}+C'$$

If I had solved the integral with the general formula I wouldn't have got partial amount of the integration constant($\dfrac{-1}{4}+C'$) as my final answer.

Simply, my question is why another constant would rise up when I do the integration with trigonometric substitution?I know that I can disregard the appeared constant but why does it even rise up?

If I'm still not clear enough, please tell me to correct my question.

Answer 1

The constant of integration is arbitrary and two anti-derivatives are equivalent if they differ by a constant.

For example, since $$ \int0\,\mathrm{d}x=C $$ we get not only $$ \int\cos(x)\,\mathrm{d}x=\sin(x)+C $$ but also $$ \begin{align} \int\cos(x)\,\mathrm{d}x &=\int(\color{#C00000}{\cos(x)}+\color{#00A000}{0})\,\mathrm{d}x\\ &=\int\color{#C00000}{\cos(x)}\,\mathrm{d}x+\int\color{#00A000}{0}\,\mathrm{d}x\\[6pt] &=\color{#C00000}{\sin(x)+C}+\color{#00A000}{C} \end{align} $$ where the constants $\color{#C00000}{C}$ and $\color{#00A000}{C}$ are possibly different constants.

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In the particular case you give, using $x=\sin(\theta)$, $$ \begin{align} \int x\,\mathrm{d}x &=\int\sin(\theta)\,\cos(\theta)\,\mathrm{d}\theta\\ &=\int\tfrac12\sin(2\theta)\,\mathrm{d}\theta\\ &=-\tfrac14\cos(2\theta)+C\tag{$\ast$}\\[3pt] &=-\tfrac14(1-2\sin^2(\theta))+C\\[3pt] &=\tfrac12\sin^2(\theta)+C-\tfrac14\\[3pt] &=\tfrac12x^2+C-\tfrac14 \end{align} $$ In step $(\ast)$, note that although $x=0$ corresponds to $\theta=0$, $-\frac14\cos(2\theta)=-\frac14$ at $\theta=0$. This is where the $-\frac14$ is introduced, if that is what you are asking about.

However, even simpler alterations to the method of integration can yield different, but equivalent, forms of the constant of integration: $$ \begin{align} \int x\,\mathrm{d}x &=\frac12\int 2x\,\mathrm{d}x\\ &=\frac12\left(x^2+C\right)\\[6pt] &=\tfrac12x^2+\tfrac12C \end{align} $$

Answer 2

There is no other constant, because you see, $\int\sin 2\theta\, d\theta = \dfrac{1 - \cos 2\theta}{2} = \sin^2 \theta$. Oh, what's that, you're saying it's just $\dfrac{-\cos 2\theta}{2}$? Well okay then, I'll say that $\int x\, dx = \dfrac{x^2}{2} -\dfrac{1}{4}$.

I hope you see what's going on. There's no unique anti-derivative. You always make a choice when you write it as some particular function. My favourite example is, $$\int \dfrac{-1}{\sqrt{1 - x^2}} dx = \cos^{-1} x$$ because $\dfrac{d}{dx} \cos^{-1}x = \dfrac{-1}{\sqrt{1 - x^2}}$, but hey $$\int \dfrac{-1}{\sqrt{1 - x^2}} dx = -\int \dfrac{1}{\sqrt{1 - x^2}} dx = -\sin^{-1} x$$ because $\dfrac{d}{dx}\sin^{-1} x = \dfrac{1}{\sqrt{1 - x^2}}$. What's going on? $\dfrac{\pi}{2} - \sin^{-1} x = \cos^{-1} x$.

The example in your question is similar, only there the constant is explicit, it is visible. But there's no reason to say that $1 - \cos2\theta$ is any less of a function that $-\cos2\theta$, so you cannot fundamentally discriminate between the two (when deciding which one the anti-derivative "ought to be").