Definite integral using only properties

[Krushnaraj Pandya]

Homework Statement

evaluate the following definite integral with limits 0 to 1$∫log(sin(πx/2)) dx$

2. The attempt at a solution
I used $∫f(x) = ∫f(a+b-x)$ to get $I=∫log(cos(πx/2))$ with the same limits. Adding them and using $log(m)+log(n)=log(mn)$ and $2sinxcosx=sin2x$ I got $2I=∫log(sin(πx)) - ∫log2$ with same limits. The second part evaluates to $log2$, I'm stuck on how to solve the first part. I'd appreciate some help, thank you. (Note-minimal use of indefinite integrals is to be done, this is supposed to be solved by properties of definite integrals such as mentioned in my attempt)

 

[Ray Vickson]

Homework Statement

evaluate the following definite integral with limits 0 to 1$∫log(sin(πx/2)) dx$

2. The attempt at a solution
I used $∫f(x) = ∫f(a+b-x)$ to get $I=∫log(cos(πx/2))$ with the same limits. Adding them and using $log(m)+log(n)=log(mn)$ and $2sinxcosx=sin2x$ I got $2I=∫log(sin(πx)) - ∫log2$ with same limits. The second part evaluates to $log2$, I'm stuck on how to solve the first part. I'd appreciate some help, thank you. (Note-minimal use of indefinite integrals is to be done, this is supposed to be solved by properties of definite integrals such as mentioned in my attempt)

Your LaTeX is very difficult to read. To use LaTeX as it is meant to be used you should type "\log" instead of "log", "\sin" instead of "sin", and "\cos" instead of "cos". Furthermore, you should type "\int_0^1" to get an integral from 0 to 1. When you do all that you get $\int_0^1 \log( \sin(\pi x /2)) \, dx$, as well as $\log(m)+\log(n) = \log(mn)$ and $2 \sin x\, \cos x = \sin 2x$, etc.

 

[Krushnaraj Pandya]

Your LaTeX is very difficult to read. To use LaTeX as it is meant to be used you should type "\log" instead of "log", "\sin" instead of "sin", and "\cos" instead of "cos". Furthermore, you should type "\int_0^1" to get an integral from 0 to 1. When you do all that you get $\int_0^1 \log( \sin(\pi x /2)) \, dx$, as well as $\log(m)+\log(n) = \log(mn)$ and $2 \sin x\, \cos x = \sin 2x$, etc.

Got it! will keep that in mind from next time, still a major improvement from normal text though- but it'll get better once I follow this. About the question though, any insights?

 

[Stephen Tashi]

You didn't explain your work clearly. Let's say the problem is to evaluate $I_0 = \int_0^1 \log( \sin(\frac{\pi}{2})) dx$ and you have it as $I_0 =$ (some stuff) $+ I_1$ where $I_1 = \int_0^1 \log(\sin(\pi x)) dx$.

Try expressing $I_1$ in terms of $I_0$.

The area under the graph of $f_1(x) = \log(\sin(\pi x))$ from $x = 0$ to $x = 1$ is twice the area under the graph of $f_1$ from $x = 0$ to $x = 1/2$. The graph of $f_1$ from $x = 0$ to $x = 1/2$ is like a contracted version of the graph of $f_0(x) = \log(\sin(\frac{\pi}{2} x))$ from $x = 0$ to $x = 1$.

 

[Ray Vickson]

Got it! will keep that in mind from next time, still a major improvement from normal text though- but it'll get better once I follow this. About the question though, any insights?

Yes, but Stephen Tashi beat me to it.

 

[Krushnaraj Pandya]

You didn't explain your work clearly. Let's say the problem is to evaluate $I_0 = \int_0^1 \log( \sin(\frac{\pi x}{2})) dx$ and you have it as $I_0 =$ (some stuff) $+ I_1$ where $I_1 = \int_0^1 \log(\sin(\pi x)) dx$.

Try expressing $I_1$ in terms of $I_0$.

The area under the graph of $f_1(x) = \log(\sin(\pi x))$ from $x = 0$ to $x = 1$ is twice the area under the graph of $f_1$ from $x = 0$ to $x = 1/2$. The graph of $f_1$ from $x = 0$ to $x = 1/2$ is like a contracted version of the graph of $f_0(x) = \log(\sin(\frac{\pi}{2} x))$ from $x = 0$ to $x = 1$.

So, trying to express $I_1 = \int_0^1 \log(\sin(\pi x)) dx$ in terms of $I_0 = \int_0^1 \log( \sin(\frac{\pi x}{2})) dx$. I put x=t/2, in $I_1$therefore dx=dt/2 and limits change from 0 to 2. while I write 1/2 outside (Then writing t as x since changing variable doesn't affect the integral-I feel like I might be wrong here??) I get $\frac{1}{2} \int_0^2 \log(\sin(\frac{\pi}{2} t))$ is equal to $\frac{1}{2} \int_0^2 \log(\sin(\frac{\pi}{2} x))$ I wrote using property-if $f(x) = f(2a - x)$ where upper limit is 2a and lower limit is 0, then it can be written as twice of that integral with half the upper limit- this evaluates to $I_1 = \int_0^1 \log( \sin(\frac{\pi x}{2})) dx$ which is equal to $I_0$ but we have $\log 2$ in RHS...so I get a wrong equation

 

[Stephen Tashi]

.so I get a wrong equation

What equation are you talking about?

$2I_0 = I_0 - \ln(2)$ is solvable

 

[Krushnaraj Pandya]

What equation are you talking about?

$2I_0 = I_0 - \ln(2)$ is solvable

Ohh right! I totally forgot there was 2 in the LHS, sorry...thanks a lot