Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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) ?

share|improve this question
    
Using complex exponential? –  peterwhy Nov 22 '13 at 6:24
    
3  
What is your definition of sine and cosine? –  David H Nov 22 '13 at 6:26
    
$\int \sin(x)dx = \mathrm{Im}(\int e^{ix}dx)$ –  Bitrex Nov 22 '13 at 6:29
6  
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 $$\lim_{y \rightarrow 0} \frac{y}{\sin y} = 1.$$ –  Noam D. Elkies Nov 22 '13 at 6:29
add comment

5 Answers 5

up vote 12 down vote accepted

As pointed out in the comment, using Weierstrass substitution,

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

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

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

share|improve this answer
    
If $x=2\arctan t$ then for calculating $dx$ you must use $\sin'(x)=\cos(x)$ –  Peter Nov 22 '13 at 6:44
1  
@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 –  lab bhattacharjee Nov 22 '13 at 6:48
add comment

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

share|improve this answer
add comment

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

share|improve this answer
add comment

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 $$\sec'x=\lim_{h\to0}\frac{\sec(x+h)-\sec x}{h}=\lim_{h\to0}\frac{\sin\frac{h}{2}\sin\frac{2x+h}{2}}{\frac{h}{2}\cos x\cos(x+h)}=\tan x\sec x.$$

share|improve this answer
add comment

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 $cos(x)$

share|improve this answer
2  
That is not the definition of the indefinite integral. –  Jp McCarthy Nov 22 '13 at 11:13
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.