WolframAlpha gives me the solution that the integral $$\int \sin^2(\sin(x)) \cot(x) \, dx = \frac12 \log(\sin(x)) - \frac12\text{Ci}(2\sin x)$$ plus a constant, where $$\text{Ci}(y) = - \int_{y}^{\infty} \frac{\cos(t)}{t} \, dt.$$ It doesn't give me any sort of derivation though so I'd appreciate any formal proof showing this equivalence.
$\begingroup$
$\endgroup$
1
-
$\begingroup$ Try substitution; sin(x)=t. $\endgroup$– KaranFeb 4 at 23:37
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
HINT
Rearrange the integrand in order to obtain the following relation:
\begin{align*} \int\sin^{2}(\sin(x))\cot(x)\mathrm{d}x & = \int\frac{\sin^{2}(\sin(x))}{\sin(x)}\mathrm{d}(\sin(x))\\\\ & = \int\frac{\sin^{2}(u)}{u}\mathrm{d}u\\\\ & = \int\frac{1 - \cos(2u)}{2u}\mathrm{d}u\\\\ & = \int\frac{\mathrm{d}u}{2u} - \int\frac{\cos(2u)}{2u}\mathrm{d}u \end{align*}
Can you take it from here?
-
$\begingroup$ Thank you; got it. Forgot my trig identities. $\endgroup$ Feb 5 at 0:22