Help solving a fraction integral

[protivakid]

Homework Statement

$$\int$$(5^{sin x}cosxdx)/(2+5^2sinx)

The Attempt at a Solution

I set u = sinx and du = cosxdx which gives me...

$$\int$$(5^udu)/(2+5^2u). I just need a little push as where to go from here, not the entire solution, just a push as what to do next. I have a feeling I need to use ln and know that a^r = r ln(a) but don't know if I am supposed to use that on the 5^u or not, any help is appreciated. Thanks in advance. :)

 

[sutupidmath]

U could take this subs, at the very beginning

$$\int\frac{5^{sin(x)}cosxdx}{2+(5^{sinx})^2}$$ so taking another substitution here will work quite nicely..

$$5^{sinx}=\sqrt2 u=>\sqrt2 du=5^{sinx}cos(x)ln(5)dx$$ now i guess you know how things turn out to be, right?

 

[protivakid]

That helped a lot, would the final answer then be (1/ln(5)$$\sqrt{}2$$)tan^-1(5^sinx/$$\sqrt{}2$$) ? The only thing I was not sure about was the ln(5) as far as how to deal with that.

 

[sutupidmath]

take the derivative of your final answer and see if you get the integrand, then its okay, because i won't bother to do the whole thing. The general format of it looks okay, but i didn't check the details.

 

[HallsofIvy]

Homework Statement

$$\int$$(5^{sin x}cosxdx)/(2+5^2sinx)

The Attempt at a Solution

I set u = sinx and du = cosxdx which gives me...

$$\int$$(5^udu)/(2+5^2u). I just need a little push as where to go from here, not the entire solution, just a push as what to do next. I have a feeling I need to use ln and know that a^r = r ln(a) but don't know if I am supposed to use that on the 5^u or not, any help is appreciated. Thanks in advance. :)

5^2u= (5^u)² so the next step would be the substitution v= 5^u, dv= ln(5) 5^udu.