Checking my Integral Proof

Checking my Integral Proof

Postby Guest » Tue Aug 18, 2020 3:48 pm

Compute [tex]\int \sin(x) \left( \frac{1}{\cos(x) + \sin(x)} + \frac{1}{\cos(x) - \sin(x)} \right)\,dx[/tex]

First, simplify the two fractions into one using a common denominator [tex]\int\sin(x)\left(\frac{1}{\cos(x)+\sin(x)}+\frac1{\cos(x)-\sin(x)}\right)dx=\int\sin(x)\left(\frac{\cos(x)-\sin(x)}{\cos^2(x)-\sin^2(x)}+\frac{\cos(x)+\sin(x)}{\cos^2(x)-\sin^2(x)}\right)dx[/tex][tex]=\int\frac{2\sin(x)\cos(x)}{\cos(2x)}dx \\
=\int\frac{\sin(2x)}{\cos(2x)}dx[/tex]
Let $u = 2x$. Taking the derivative, $\frac{du}{dx} = 2,$ or $dx = \frac{du}{2}.$
Plugging in $u$ for $2x,$ and $\frac{du}{2}$ for $dx,$ we get $\frac{1}{2}\int \frac{\sin(u)}{\cos(u)}\,du.$
Taking the integral of $\tan(u)$, as solved in problem 10, we get $-\log|\cos(u)|+C.$
Therefore, $\frac{1}{2}\int \frac{\sin(u)}{\cos(u)}\,du = -\frac{1}{2}\log|\cos(u)|+C.$
Finally, subbing $2x$ for $u$, we get $\boxed{-\frac{1}{2}\log|\cos(2x)|+C}$ where $C$ is a constant
Guest
 

Re: Checking my Integral Proof

Postby mario99 » Wed Aug 19, 2020 5:17 am

Perfectly, correct! 8)

mario99
 
Posts: 3
Joined: Wed Aug 19, 2020 4:45 am
Reputation: 0

Re: Checking my Integral Proof

Postby HallsofIvy » Fri Aug 28, 2020 6:34 pm

Rather than using [tex]\frac{sin(x)}{cos(x)}= tan(x)[/tex] to integrate [tex]\int \frac{sin(x)}{cos(x)}dx[/tex], I would let u= cos(x) so that du= -sin(x)dx. The integral becomes [tex]\int \frac{sin(x)}{cos(x)} dx= -\int \frac{1}{u}du= -ln(u)+ C= -ln(cos(x))+ C[/tex]

HallsofIvy
 
Posts: 341
Joined: Sat Mar 02, 2019 9:45 am
Reputation: 119


Return to Calculus - integrals, lim, functions



Who is online

Users browsing this forum: No registered users and 2 guests