Question

int_{0.1}^{0.7}cosx ln(2sinx)dx

Answer 1

\[\int\limits_{0.1}^{0.7}cosx \ln(2sinx)dx\]

Answer 2

Let \(u=2\sin x\), so \(\dfrac{1}{2}~du=\cos x~dx\).

Answer 3

Integral changes to
\[\frac{1}{2}\int_{2\sin(0.1)}^{2\sin(0.7)}\ln u~du\]
Then integrate by parts, or use the formula for \(\int\ln u~du\) if you know it.

Answer 4

could you please explain more

Answer 5

As in continue? or explain what I did so far?

Answer 6

continue

Answer 7

\[\frac{1}{2}\int_{2\sin(0.1)}^{2\sin(0.7)}\ln u~du=\frac{1}{2}\left[fg-\int g~df\right]_{2\sin(0.1)}^{2\sin(0.7)}\]
where you have
\[\begin{matrix}f=\ln u&&&dg=du\\df=\frac{du}{u}&&&g=u\end{matrix}\]
\[\frac{1}{2}\left[fg-\int g~df\right]_{2\sin(0.1)}^{2\sin(0.7)}=\frac{1}{2}\left[\left[u\ln u\right]_{2\sin(0.1)}^{2\sin(0.7)}-\int_{2\sin(0.1)}^{2\sin(0.7)}u\left(\frac{du}{u}\right)\right]\]
Simplified a bit, you have
\[\frac{1}{2}\left[\left[u\ln u\right]_{2\sin(0.1)}^{2\sin(0.7)}-\int_{2\sin(0.1)}^{2\sin(0.7)}du\right]\\
\frac{1}{2}\Bigg[\left[u\ln u\right]_{2\sin(0.1)}^{2\sin(0.7)}-\left[u\right]_{2\sin(0.1)}^{2\sin(0.7)}\Bigg]\]

Answer 8

so what you get for answer

Answer 9

just checking