8
$\begingroup$

I can prove that

$$\int\sin(x)dx=-\cos(x)+C$$

by using $\cos'(x)=-\sin(x)$ and $\sin'(x)=\cos(x)$. Are there other proofs not involving this (at least, not explicitly) ?

$\endgroup$
  • $\begingroup$ Using complex exponential? $\endgroup$ – peterwhy Nov 22 '13 at 6:24
  • $\begingroup$ Try en.wikipedia.org/wiki/Tangent_half-angle_substitution $\endgroup$ – lab bhattacharjee Nov 22 '13 at 6:24
  • 3
    $\begingroup$ What is your definition of sine and cosine? $\endgroup$ – David H Nov 22 '13 at 6:26
  • $\begingroup$ $\int \sin(x)dx = \mathrm{Im}(\int e^{ix}dx)$ $\endgroup$ – Bitrex Nov 22 '13 at 6:29
  • 7
    $\begingroup$ It's possible (though cumbersome) to evaluate $\int_{x_1}^{x_2} \sin x \, dx$ as a limit of Riemann sums via the formula for $\sum_{i=1}^n \sin(x_1 + i y)$; at the end you'll need to use $$\lim_{y \rightarrow 0} \frac{y}{\sin y} = 1.$$ $\endgroup$ – Noam D. Elkies Nov 22 '13 at 6:29
0
$\begingroup$

Let $x=\sec^{-1}u$ then we have $$\sin(\sec^{-1}u)=\frac{\sqrt{u^2-1}}{|u|},dx=\frac{du}{|u|\sqrt{u^2-1}}$$ therefore $$\int\sin x\,dx=\int\frac{du}{u^2}=\frac{-1}{u}=-\cos x.$$ Note that $$\sec'x=\lim_{h\to0}\frac{\sec(x+h)-\sec x}{h}=\lim_{h\to0}\frac{\sin\frac{h}{2}\sin\frac{2x+h}{2}}{\frac{h}{2}\cos x\cos(x+h)}=\tan x\sec x.$$

| cite | improve this answer | |
$\endgroup$
12
$\begingroup$

Just use the taylor series and integrate term by term, you recognise the new Taylor series as $-\cos x$

| cite | improve this answer | |
$\endgroup$
10
$\begingroup$

Absolutely. You can integrate the exponential form $$\sin(x) = \frac{e^{ix} - e^{-ix}}{2i} ,$$ and then return that result back into your desired integral.

Note that $$\cos(x) = \frac{e^{ix} + e^{-ix}}{2}.$$

| cite | improve this answer | |
$\endgroup$
10
$\begingroup$

As pointed out in the comment, using Weierstrass substitution,

$\displaystyle \tan\frac x2=t\implies \sin x=\frac{2t}{1+t^2}$ and $\displaystyle x=2\arctan t\implies dx=\frac2{1+t^2}dt$

$$\int \sin xdx=\int \frac{2t}{(1+t^2)^2}dt=\int \frac{du}{(1+u)^2}\text{ (putting }t^2=u)$$

$$=-\frac1{1+u}=-\frac1{1+t^2}=-2\cos^2\frac x2+C=C-1-\cos x$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ If $x=2\arctan t$ then for calculating $dx$ you must use $\sin'(x)=\cos(x)$ $\endgroup$ – user95733 Nov 22 '13 at 6:44
  • 2
    $\begingroup$ @user95733, why? we can use $$\frac{d (\tan y)}{dy}=\lim_{h\to0}\frac{\tan(y+h)-\tan y}h=\sec^2 y$$ as well $\endgroup$ – lab bhattacharjee Nov 22 '13 at 6:48
3
$\begingroup$

It's important to notice that the definition of indefinite integral. Let $D \subset R$ and $f:D \mapsto R$ be a function. Then the indefinite integral of $f$ is defined as a function $F: D\mapsto R$ such that $F$ is differentiable on $D$ and $F'=f$

So there's no way to prove the indefinite integral without using its definition. But of course, it's possible to write the function $cos(x)$ in different ways i.e. cos$x$=$\frac{(e^{ix}+e^{-ix})}{2}$ or other ways and show the expressions are equal to $cos(x)$

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ That is not the definition of the indefinite integral. $\endgroup$ – JP McCarthy Nov 22 '13 at 11:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy