MHB Integral Evaluation: x^2+4 & 2+2sinx+cosx

[mathworker]

Evaluate integrals:
$$1) \int\frac{dx}{x^2\sqrt{x^2+4}}$$
$$2) \int\frac{dx}{2+2\text{sin}x+\text{cos}x}$$

 

[MarkFL]

1.)

Spoiler

I would begin with the substitution:

$$x=2\tan(\theta)\,\therefore\,dx=2\sec^2(\theta)$$

Thus, we now have:

$$\frac{1}{4}\int\frac{\cos(\theta)}{\sin^2(\theta)}\,d\theta$$

Now, let:

$$u=\sin(\theta)\,\therefore\,du=\cos(\theta)\,d \theta$$

and we now have:

$$\frac{1}{4}\int u^{-2}\,du=-\frac{1}{4u}+C$$

Back-substitute for $u$:

$$-\frac{1}{4\sin(\theta)}+C$$

Back-substitute for $\theta$:

$$\int\frac{dx}{x^2\sqrt{x^2+4}}=-\frac{\sqrt{x^2+4}}{4x}+C$$

 

[MarkFL]

2.)

Spoiler

I would begin by rewriting the denominator of the integrand:

$$2+2\sin(x)+\cos(x)=2\left(\sin^2\left(\frac{x}{2} \right)+\cos^2\left(\frac{x}{2} \right) \right)+4\sin\left(\frac{x}{2} \right)\cos\left(\frac{x}{2} \right)+\cos^2\left(\frac{x}{2} \right)-\sin^2\left(\frac{x}{2} \right)=$$

$$\left(\sin\left(\frac{x}{2} \right)+\cos\left(\frac{x}{2} \right) \right)\left(\sin\left(\frac{x}{2} \right)+3\cos\left(\frac{x}{2} \right) \right)$$

Next, the numerator may be rewritten as:

$$1=\left(\frac{1}{2}\cos\left(\frac{x}{2} \right)-\frac{1}{2}\sin\left(\frac{x}{2} \right) \right)\left(\sin\left(\frac{x}{2} \right)+3\cos\left(\frac{x}{2} \right) \right)-\left(\frac{1}{2}\cos\left(\frac{x}{2} \right)-\frac{3}{2}\sin\left(\frac{x}{2} \right) \right)\left(\sin\left(\frac{x}{2} \right)+\cos\left(\frac{x}{2} \right) \right)$$

Thus, the integrand may be rewritten as:

$$\frac{\frac{1}{2}\cos\left(\frac{x}{2} \right)-\frac{1}{2}\sin\left(\frac{x}{2} \right)}{\sin\left(\frac{x}{2} \right)+\cos\left(\frac{x}{2} \right)}-\frac{\frac{1}{2}\cos\left(\frac{x}{2} \right)-\frac{3}{2}\sin\left(\frac{x}{2} \right)}{\sin\left(\frac{x}{2} \right)+3\cos\left(\frac{x}{2} \right)}$$

Hence:

$$\int\frac{dx}{2+2\sin(x)+\cos(x)}=\ln\left|\frac{ \sin\left(\frac{x}{2} \right)+\cos\left(\frac{x}{2} \right)}{ \sin\left(\frac{x}{2} \right)+3\cos\left(\frac{x}{2} \right)} \right|+C$$