#I = int (sin3x)/sqrt(5+cos3x)dx#
Generally, when we have integrals where two trigonometric functions are being divided, we should tend to substitute them such that the numerator or denominator cancels out.
Firstly, let #u = 5+cos3x#. Then,
#(du)/dx = d/dx [5+cos3x]=d/dxcos3x#
We have to use the #color(red)("chain rule")#, which states that:
#[f(g(x))]' = f'(g(x))*g'(x)#
In our case, #f(x) = cosx# and #g(x) = 3x#.
#:. [cos3x]' = -3sin3x#
#d/dx cos3x = -3sin3x#
We have:
#du = -3sin3x dx => dx=-1/(3sin3x)du#
By substituting #u# into the integral, we have:
#int (sin3x)/sqrt(5+cos3x)dx= int (sin3x)/sqrtu* (du)/(-3sin3x)#
You can see that the #sin3x# cancels, as we needed.
#=> I = -1/3int cancel(sin3x)/sqrtu* (du)/cancel(sin3x) = -1/3 int 1/sqrtu du#
Since #1/sqrtu = u^(-1/2)#, we get:
#I=-1/3int 1/sqrtu = -1/3 intu^(-1/2)du#
We know that, for any #n !=-1#:
#int x^n dx = x^(n+1)/(n+1)+ C#
Thus, our integral is equal to:
#I = -1/3 * u^(1-1/2)/(1-1/2)+ C = -2/3 sqrtu + C#
Knowing that #u = 5+cos3x#, we finally get our answer:
#I = -2/3 sqrt(5+cos3x)+ C#
