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!!
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!!
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.
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.