Question

Find ∫(cos√x) dx=?

Answer 1

Let's solve the integral $\int \cos(\sqrt{x}) \, dx$. 

1. Substitution Let $u = \sqrt{x}$. Then $x = u^2$, and $dx = 2u \, du$. 
Substituting these into the integral, we have: $\int \cos(\sqrt{x}) \, dx = \int \cos(u) \cdot 2u \, du = 2 \int u \cos(u) \, du$ 

2. Integration by Parts Now, we need to use integration by parts to solve $\int u \cos(u) \, du$. Recall the integration by parts formula: $\int v \, dw = vw - \int w \, dv$ 
Let $v = u$ and $dw = \cos(u) \, du$. Then $dv = du$ and $w = \sin(u)$. 
Applying the formula: $\int u \cos(u) \, du = u \sin(u) - \int \sin(u) \, du$ 
$\int u \cos(u) \, du = u \sin(u) - (-\cos(u)) + C_1$ 
$\int u \cos(u) \, du = u \sin(u) + \cos(u) + C_1$ 

3. Substitute Back Now, substitute this result back into our integral and then substitute $u = \sqrt{x}$: $2 \int u \cos(u) \, du = 2(u \sin(u) + \cos(u)) + C$ 
$2 \int u \cos(u) \, du = 2(\sqrt{x} \sin(\sqrt{x}) + \cos(\sqrt{x})) + C$ 

Final Answer Therefore, the integral is: $\int \cos(\sqrt{x}) \, dx = 2\sqrt{x} \sin(\sqrt{x}) + 2\cos(\sqrt{x}) + C$