To evaluate the indefinite integral
[tex]\[
\int 5 e^{5 v} \sin \left( e^{5 v} \right) \, d v,
\][/tex]
we begin by considering a substitution that simplifies the integrand.
1. Let [tex]\( u = e^{5v} \)[/tex]. This substitution simplifies the exponential term inside the sine function.
2. Next, we find the differential [tex]\( du \)[/tex] in terms of [tex]\( dv \)[/tex]:
[tex]\[
du = \frac{d}{dv}\left(e^{5v}\right) \, dv = 5e^{5v} \, dv.
\][/tex]
3. This can be rearranged to solve for [tex]\( dv \)[/tex]:
[tex]\[
dv = \frac{du}{5e^{5v}}.
\][/tex]
4. Substitute [tex]\( u \)[/tex] and [tex]\( dv \)[/tex] back into the integral:
[tex]\[
\int 5 e^{5v} \sin \left(e^{5v}\right) \, dv = \int 5 e^{5v} \sin \left(u \right) \left( \frac{du}{5e^{5v}} \right) = \int \sin (u) \, du.
\][/tex]
5. The integral of [tex]\( \sin(u) \)[/tex] with respect to [tex]\( u \)[/tex] is:
[tex]\[
\int \sin(u) \, du = -\cos(u) + C,
\][/tex]
where [tex]\( C \)[/tex] is the constant of integration.
6. Finally, revert the substitution [tex]\( u = e^{5v} \)[/tex] back to the original variable [tex]\( v \)[/tex]:
[tex]\[
-\cos(u) + C = -\cos \left(e^{5v}\right) + C.
\][/tex]
Therefore, the result of the indefinite integral is:
[tex]\[
\int 5 e^{5 v} \sin \left( e^{5 v} \right) \, d v = -\cos \left( e^{5 v} \right) + C.
\][/tex]
