what are the possible answers we can get for the below intergral?

Question

Could you please tell me what are the possible answers (if there is more than one) for the following indefinite integral?

$$\int \dfrac{\cos(\sqrt{6x})}{\sqrt{6x}}dx$$

Answer 1

Set $6x=y^2$. We then have $6dx = 2ydy$. Hence, $$\int \dfrac{\cos(\sqrt{6x})}{\sqrt{6x}}dx = \int \dfrac{\cos(y)}{y} \dfrac{ydy}3 = \dfrac13 \int \cos(y)dy = \dfrac{\sin(\sqrt{6x})}3 + \text{constant}$$

Answer 2

The double appearance of $\sqrt{6x}$ should suggest at least trying the substitution $u=\sqrt{6x}$. (With more experience you’ll know immediately that it’s a good idea, because you’ll be able to look ahead and see that $du$ will the $\sqrt{6x}$ in the denominator that you need.) Now you have $u=(6x)^{1/2}$, so $$du=\frac12(6x)^{-1/2}\cdot6\,dx=\frac3{\sqrt{6x}}dx\;.$$ This is perfect: except for the factor of $3$, it matches the $\dfrac{dx}{\sqrt{6x}}$ in your integral, and constant factors are never a problem.

That is, you almost have $\int\cos u\,du$; with a factor of $3$ in the integral you would have $\int\cos u\,du$, so write

$$\int\frac{\cos\sqrt{6x}}{\sqrt{6x}}dx=\frac13\int\cos\sqrt{6x}\cdot\frac3{\sqrt{6x}}dx\;,$$

make the substitution, and you have

$$\frac13\int\cos u\,du\;,$$

which readily integrates to $\dfrac13\sin u+C$. Now just reverse the substitution, and you’re home free:

$$\int\frac{\cos\sqrt{6x}}{\sqrt{6x}}dx=\frac13\sin\sqrt{6x}+C\;.$$