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
