2
$\begingroup$

How do I integrate the indefinite integral, $\frac {\sin t}{t+1}$ w.r.t to t.

$$\int \frac{\sin t}{t+1} \,dt$$

I've tried by parts, but seems impossible. I can't think of a good subsitution too.

Help appreciated!!

$\endgroup$
3
  • $\begingroup$ Here is some tutorial about writing integrals and we also have tutorial for writing mathematics in general. $\endgroup$ Commented Oct 27, 2012 at 7:23
  • $\begingroup$ Pretty sure that isn't going to be an elementary function. $\endgroup$
    – Mike
    Commented Oct 27, 2012 at 7:56
  • $\begingroup$ Thanks martin XD I was on a iphone so i was going to find it hard to google how to write integrals. $\endgroup$ Commented Oct 27, 2012 at 14:38

2 Answers 2

7
$\begingroup$

I don't think there is an elementary integral, but we can use special functions.

Use the substitution $s=t+1$ giving $$ \begin{align} \int\frac{\sin(t)}{t+1}\,\mathrm{d}t &=\int\frac{\sin(s-1)}{s}\,\mathrm{d}s\\ &=\int\frac{\sin(s)\cos(1)-\cos(s)\sin(1)}{s}\,\mathrm{d}s\\ &=\cos(1)\mathrm{Si}(s)-\sin(1)\mathrm{Ci}(s)+C\\[6pt] &=\cos(1)\mathrm{Si}(t+1)-\sin(1)\mathrm{Ci}(t+1)+C \end{align} $$ Where $\mathrm{Si}(x)$ is the Sine Integral and $\mathrm{Ci}(x)$ is the Cosine Integral.

$\endgroup$
1
  • 1
    $\begingroup$ This is the only way, I think. $\endgroup$
    – Mikasa
    Commented Oct 27, 2012 at 8:09
0
$\begingroup$

Another approach (an electrician's point of view):

Consider the parameter-dependent integral:

$$I(a)=\int \frac{\sin at}{t+1} \,dt$$ The Laplace transform of $I(a)$:

$$\mathcal{L}\,I(a)=\frac{\ln(t^2+s^2)}{2(1+s^2)}-\frac{\ln(t+1)}{1+s^2}+\frac{s\arctan{\frac{t}{s}}}{1+s^2}=F(s) $$ The inverse Laplace transform:

$$\mathcal{L^{-1}}F(s)=I(a)=-(\sin a) \ln(t+1)+\int_{0}^{a}\frac{\sin[(t+1)x-a]}{x}dx+\text{C}$$ I would say that the result is even a closed form solution.

$\endgroup$
4
  • $\begingroup$ How is the result any more 'closed-form' than the starting point was? $\endgroup$ Commented Oct 27, 2012 at 14:46
  • $\begingroup$ Haven't learnt laplace yet, but thanks anyway!!! Seems like its impossible to integrate with beginning calculus. $\endgroup$ Commented Oct 27, 2012 at 14:46
  • $\begingroup$ @StevenStadnicki It does not include an indefinite integral. $\endgroup$ Commented Oct 27, 2012 at 14:52
  • $\begingroup$ @MartinGales But it includes an arbitrary constant of integration! I might as well say '$\int \frac{\sin t}{t+1} dt = \int_0^t\frac{\sin x}{x+1} dx+C$'. It's true, and it expresses the integral in terms of a definite integral, but it's useless for evaluating the initial integral and I wouldn't call it any more closed-form. $\endgroup$ Commented Oct 27, 2012 at 15:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .