Proofs and integration

[Erik Lehnsherr]

Salutations,

I don't understand the concept of mathematical proof. This is an issue in my attempts at solving these problems. Along with mathematical proof, I need some assistance in solving these problems, please. I could also fare well with some hints, please.

1.) Show that \(\displaystyle \int x^n \sin x\,dx = -x^n \cos x + n \int x^{n-1} \cos x\,dx.\)

2.) Use the substitution \(\displaystyle t = \tan\theta\) to show that \(\displaystyle \int \frac{1}{2\cos^2\theta - 1} d\theta = \int \frac{1}{1 - t^2} dt.\)

 

[tkhunny]

These are not proofs. These are mere demonstrations.

1) Try integration by parts.

2) Do the substitution.

 

[soroban]

Hello, Erik Lehnsherr!

\(\displaystyle \text{2) Use the substitution }t = \tan\theta\,\text{ to show that: }\;\int \frac{1}{2\cos^2\theta - 1}\,d\theta \;=\; \int \frac{1}{1 - t^2}\,dt\)

\(\displaystyle \text{Let: }\,\tan\theta \,=\,t \quad\Rightarrow\quad \theta \,=\,\arctan t \quad\Rightarrow\quad d\theta \:=\:\frac{dt}{1+t^2}\)

\(\displaystyle \text{Since }\,\tan\theta \:=\:\frac{t}{1} \:=\:\frac{opp}{adj},\:\text{ Pythagorus says: }\,hyp \:=\:\sqrt{1+t^2}\)
. . . . \(\displaystyle \text{Hence: }\:\cos\theta \:=\:\frac{1}{\sqrt{1+t^2}}\)


\(\displaystyle \text{Substitute: }\;\int\frac{1}{2\cos^2\theta-1}\,d\theta \;=\;\int\frac{1}{2\left(\frac{1}{\sqrt{1+t^2}}\right)^2 - 1}\,\cdot\,\frac{dt}{1+t^2} \;=\;\int\frac{1}{\frac{2}{1+t^2} - 1}\,\cdot\,\frac{dt}{1+t^2}\)

. . . . . . . . . . \(\displaystyle =\;\int\frac{1}{2-(1+t^2)}\,dt \;=\;\int\frac{1}{1-t^2}\,dt\)

 

[Erik Lehnsherr]

1.) Show that \(\displaystyle \int x^n \sin x\,dx = -x^n \cos x + n \int x^{n-1} \cos x\,dx.\)

\(\displaystyle \int x^n \sin x\, dx\)

\(\displaystyle v = x^n\,\,\,\,\,\,\,\,\,\,\frac{du}{dx} = \sin x\)

\(\displaystyle \frac{dv}{dx} = nx^{n - 1}\,\,\,\,\,\,\,\,\,u = - \cos x\)


\(\displaystyle \int v\frac{du}{dx} dx\, =\, uv - \int u \frac{dv}{dx} dx\)

\(\displaystyle \int x^n \sin x\, dx = (- \cos x)(x^n) - \int (- \cos x)(nx^{n-1})\, dx\)
\(\displaystyle = -\cos x\,\,x^n + n \int x^{n-1}\cos x\, dx\)
\(\displaystyle = - x^n\,\cos x + n \int x^{n-1}\cos x\,dx\)
As desired.

 

[Erik Lehnsherr]

Thank you tkhunny and soroban.

 

[Erik Lehnsherr]

Hello, Erik Lehnsherr!

\(\displaystyle \text{2) Use the substitution }t = \tan\theta\,\text{ to show that: }\;\int \frac{1}{2\cos^2\theta - 1}\,d\theta \;=\; \int \frac{1}{1 - t^2}\,dt\)

\(\displaystyle \text{Let: }\,\tan\theta \,=\,t \quad\Rightarrow\quad \theta \,=\,\arctan t \quad\Rightarrow\quad d\theta \:=\:\frac{dt}{1+t^2}\)

\(\displaystyle \text{Since }\,\tan\theta \:=\:\frac{t}{1} \:=\:\frac{opp}{adj},\:\text{ Pythagorus says: }\,hyp \:=\:\sqrt{1+t^2}\)
. . . . \(\displaystyle \text{Hence: }\:\cos\theta \:=\:\frac{1}{\sqrt{1+t^2}}\)


\(\displaystyle \text{Substitute: }\;\int\frac{1}{2\cos^2\theta-1}\,d\theta \;=\;\int\frac{1}{2\left(\frac{1}{\sqrt{1+t^2}}\right)^2 - 1}\,\cdot\,\frac{dt}{1+t^2} \;=\;\int\frac{1}{\frac{2}{1+t^2} - 1}\,\cdot\,\frac{dt}{1+t^2}\)

. . . . . . . . . . \(\displaystyle =\;\int\frac{1}{2-(1+t^2)}\,dt \;=\;\int\frac{1}{1-t^2}\,dt\)

I believe the second part of the question is a follow-up. This is my best attempt. Is it correct?

\(\displaystyle \text{Part b.) Show that}\,\,\int_0^\frac{\pi}{6} \frac{1}{2\cos^2\,\theta-1}=\frac{1}{2}\ln(2 + \sqrt3)\)

\(\displaystyle \int \frac{1}{2\cos^2 \theta - 1} = \frac{1}{2} (\log(\sin(\theta) + \cos(\theta)) - \log(\cos(\theta) - \sin(\theta))) + K\)

\(\displaystyle \text{So:}\,\,\,\frac{1}{2} (\log(\sin(\theta) + \cos(\theta)) - \log(\cos(\theta) - \sin(\theta))) + K\)

\(\displaystyle \text{In the following I used}\) \(\displaystyle F|_a^b\) \(\displaystyle \Rightarrow F |_0^\frac{\pi}{6}\)

\(\displaystyle \frac{1}{2} \bigg(\log\bigg(\sin\bigg(\frac{\pi}{6}\bigg) + \cos\bigg(\frac{\pi}{6}\bigg)\bigg) - \log\bigg(\cos\bigg(\frac{\pi}{6}\bigg) - \sin\bigg(\frac{\pi}{6}\bigg)\bigg)\bigg)\bigg -\)
\(\displaystyle \frac{1}{2} \bigg(\log\bigg(\sin\bigg(0\bigg) + \cos\bigg(0\bigg)\bigg) - \log\bigg(\cos\bigg(0\bigg) - \sin\bigg(0\bigg)\bigg)\bigg)\)


\(\displaystyle = \frac{1}{2}\log\bigg(\frac{1}{2} + \frac{\sqrt3}{2} - \log\bigg(\frac{\sqrt3}{2} - \frac{1}{2}\bigg)\bigg)\)

 

[BigGlenntheHeavy]

\(\displaystyle For \ part \ B, \ use \ identity \ 2cos^2(x)-1 \ = \ cos(2x), \ and \ then \ let \ u \ = \ 2x.\)

 

[Erik Lehnsherr]

\(\displaystyle For \ part \ B, \ use \ identity \ 2cos^2(x)-1 \ = \ cos(2x), \ and \ then \ let \ u \ = \ 2x.\)

I got the same thing:

\(\displaystyle \text{Part b.) Show that}\,\,\int_0^\frac{\pi}{6} \frac{1}{2\cos^2\,\theta-1}=\frac{1}{2}\ln(2 + \sqrt3)\)

\(\displaystyle 2\cos^2 \theta - 1 \equiv \cos(\theta)\)

\(\displaystyle \int \frac{1}{\cos\theta} = \frac{1}{2} (\log(\sin(\theta) + \cos(\theta)) - \log(\cos(\theta) - \sin(\theta))) + K\)

\(\displaystyle \text{So:}\,\,\,\frac{1}{2} (\log(\sin(\theta) + \cos(\theta)) - \log(\cos(\theta) - \sin(\theta))) + K\)

\(\displaystyle \text{In the following I used}\) \(\displaystyle F|_a^b\) \(\displaystyle \Rightarrow F |_0^\frac{\pi}{6}\)

\(\displaystyle \frac{1}{2} \bigg(\log\bigg(\sin\bigg(\frac{\pi}{6}\bigg) + \cos\bigg(\frac{\pi}{6}\bigg)\bigg) - \log\bigg(\cos\bigg(\frac{\pi}{6}\bigg) - \sin\bigg(\frac{\pi}{6}\bigg)\bigg)\bigg)\bigg -\)
\(\displaystyle \frac{1}{2} \bigg(\log\bigg(\sin\bigg(0\bigg) + \cos\bigg(0\bigg)\bigg) - \log\bigg(\cos\bigg(0\bigg) - \sin\bigg(0\bigg)\bigg)\bigg)\)


\(\displaystyle = \frac{1}{2}\log\bigg(\frac{1}{2} + \frac{\sqrt3}{2} - \log\bigg(\frac{\sqrt3}{2} - \frac{1}{2}\bigg)\bigg)\)

Using an identity doesn't change anything since it's merely a substitution.
I'm terrible at this. Please, I require more help.

 

[tkhunny]

Careful, there. Using that particular identity changes things from quadratic to linear. That is definitely something.

 

[BigGlenntheHeavy]

\(\displaystyle \int_{0}^{\pi/6}\frac{d\theta}{2cos^2(\theta)-1} \ = \ \int_{0}^{\pi/6}\frac{d\theta}{cos(2\theta)}\)

\(\displaystyle Let \ u \ = \ 2\theta, \ du \ = \ 2d\theta, \ then \ \frac{1}{2}\int_{0}^{\pi/3}\frac{du}{cos(u)} \ = \ \frac{1}{2}\int_{0}^{\pi/3}sec(u)du,\)

\(\displaystyle = \ \frac{1}{2}[ln(sec|u|+tan|u|)]_{0}^{\pi/3} \ = \ \frac{1}{2}ln(2+\sqrt3)\)

\(\displaystyle tkhunny, \ where \ is \ there \ a \ problem \ using \ the \ identity?\)

\(\displaystyle An \ identity \ is \ an \ identity, \ it \ doesn't \ change \ anything.\)

 

[Erik Lehnsherr]

\(\displaystyle \int_{0}^{\pi/6}\frac{d\theta}{2cos^2(\theta)-1} \ = \ \int_{0}^{\pi/6}\frac{d\theta}{cos(2\theta)}\)

\(\displaystyle Let \ u \ = \ 2\theta, \ du \ = \ 2d\theta, \ then \ \frac{1}{2}\int_{0}^{\pi/3}\frac{du}{cos(u)} \ = \ \frac{1}{2}\int_{0}^{\pi/3}sec(u)du,\)

\(\displaystyle = \ \frac{1}{2}[ln(sec|u|+tan|u|)]_{0}^{\pi/3} \ = \ \frac{1}{2}ln(2+\sqrt3)\)

\(\displaystyle tkhunny, \ where \ is \ there \ a \ problem \ using \ the \ identity?\)

\(\displaystyle An \ identity \ is \ an \ identity, \ it \ doesn't \ change \ anything.\)

What's the reason for the change in intervals\, \(\displaystyle \frac{\pi}{6}\,to\,\frac{\pi}{3}?\)
My guess is the part where you factored out the constant to get: \(\displaystyle \frac{1}{2}\int_{0}^{\pi/3}\frac{du}{cos(u)}\)

 

[galactus]

When you make a substitution on a definite integral, you must change the limits of integration, accordingly.

Since \(\displaystyle u=2{\theta}\), then \(\displaystyle u=2(\frac{\pi}{6})=\frac{\pi}{3}\)

 

[Erik Lehnsherr]

Many thanks BigGlenntheHeavy and galactus!