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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!!

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    Thanks martin XD I was on a iphone so i was going to find it hard to google how to write integrals.2012-10-27

2 Answers 2

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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.

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    This is the only way, I think.2012-10-27
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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.

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    @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.2012-10-27