Question

Evaluate the following integrals: $$ \int \frac{\sin x d x}{\sqrt{\cos ^{2} x+4 \cos x+1}} $$

Short answer

Question: Evaluate the integral $\int \frac{\sin x d x}{\sqrt{\cos^2 x + 4\cos x + 1}}$. Answer: $-\operatorname{arcosh}\frac{\cos x + 2}{\sqrt{3}} + C$

Step-by-step solution

Complete the square

Let's complete the square for the expression inside the square root: $$ \cos^2 x + 4\cos x + 1 = (\cos x + 2)^2 - 3 $$ Now we can rewrite the integral as: $$ \int \frac{\sin x d x}{\sqrt{(\cos x + 2)^2 - 3}} $$

Apply substitution

Now, we make the substitution \(u = \cos x + 2\) which implies \(du = -\sin x dx\). We can also rewrite \(\sin x dx = -d u\) and the integral becomes: $$ \int \frac{- du}{\sqrt{u^2 - 3}} $$

Evaluate the integral

The integral now becomes a standard hyperbolic trigonometric substitution form. We can recognize it as the arcosh function: $$ -\int \frac{du}{\sqrt{u^2 - 3}} = -\operatorname{arcosh} \frac{u}{\sqrt{3}} + C $$ where \(C\) is the constant of integration.

Revert to the original variable

Now substitute back the original variable \(x\) using \(u = \cos x + 2\): $$ -\operatorname{arcosh} \frac{u}{\sqrt{3}} + C = -\operatorname{arcosh} \frac{\cos x + 2}{\sqrt{3}} + C $$ Finally, the evaluated integral is: $$ \int \frac{\sin x d x}{\sqrt{\cos ^{2} x+4 \cos x+1}} = -\operatorname{arcosh} \frac{\cos x + 2}{\sqrt{3}} + C $$