Welcome to our community

Be a part of something great, join today!

Evaluate Integrals

mathworker

Active member
May 31, 2013
118
Evaluate integrals:
\(\displaystyle 1) \int\frac{dx}{x^2\sqrt{x^2+4}}\)
\(\displaystyle 2) \int\frac{dx}{2+2\text{sin}x+\text{cos}x}\)
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
1.)
I would begin with the substitution:

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

Thus, we now have:

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

Now, let:

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

and we now have:

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

Back-substitute for $u$:

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

Back-substitute for $\theta$:

\(\displaystyle \int\frac{dx}{x^2\sqrt{x^2+4}}=-\frac{\sqrt{x^2+4}}{4x}+C\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
2.)

I would begin by rewriting the denominator of the integrand:

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

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

Next, the numerator may be rewritten as:

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

Thus, the integrand may be rewritten as:

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

Hence:

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