Integral help, Leibniz?

Question

Let β = ∫ ln ( x + α) dx. Find dβ/dα. The integral runs from sin α on the top to cos α on the bottom.

You can use B's and a's in the explaination, I don't care about the Greek stuff :P

Answer 1

As you say, it's Leibniz Rule time...

dβ/dα = [ln(sin(α) + α)]*cos(α) - [ln(cos(α) + α)]*(-sin(α)) + ∫ 1/(x + α) dx

The (definite) integral is easy and I leave it to you.

Answer 2

The indefinite integral is
∫ln(x) = xln(x) – x + c
For the interval [cosα, sinα], the definite integral is
β = [xln(sinα) – sinα + c] - [xln(cosα) – cosα + c]
β = x[(ln(sinα)–ln(cosα)] + (cosα – sinα) + (c – c)
β = xln(sinα/cosα) + cosα – sinα
β = xln(tanα) + cosα – sinα

dβ/dα = d[xln(tanα) + cosα – sinα]/dα
dβ/dα = x*d[ln(tanα)]/dα - sinα – cosα
dβ/dα = x*(1/tanα)(sec²α) - sinα – cosα
dβ/dα = x*(cosα/sinα)(1/cos²α) - sinα – cosα
dβ/dα = x*(1/sinα)(1/cosα) - sinα – cosα
dβ/dα = x*cosecα*secα - sinα – cosα