# Evaluate this index Integral containing Trig

## Homework Statement

Folks Evaluate $B_{11}$ given

$\displaystyle B_{ij}=\int_0^1 (1+x) \frac{d \phi_i}{dx} \frac{ d\phi_j}{dx} dx$ where $\phi_i= sin i \pi x$ and $\phi_j=sin j \pi x$

## The Attempt at a Solution

I calculate $\displaystyle B_{ij}=\int_0^1 (1+x)[ i \pi \cos(i \pi x))(j \pi \cos(j \pi x)]=ij \pi^2 \int_0^1 (1+x) \left[\frac{1}{2} [\cos(i+j)\pi x+\cos(i-j)\pi x\right ]dx$

$\displaystyle = \frac{ij \pi^{2}}{2} \int_0^1\left [ \cos(i+j) \pi x+\cos(i-j) \pi x +x \cos(i+j) \pi x+x \cos(i-j) \pi x \right ]dx$

Now for the second and last term in the integrand if we substitue $i=j=1$ after integrating we will get a 0 in the denominator ....but the book calculates $B_{ij}=\frac{3 \pi^2}{4}$

What have I dont wrong?

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SammyS
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## Homework Statement

Folks Evaluate $B_{11}$ given

$\displaystyle B_{ij}=\int_0^1 (1+x) \frac{d \phi_i}{dx} \frac{ d\phi_j}{dx} dx$ where $\phi_i= sin i \pi x$ and $\phi_j=sin j \pi x$

## The Attempt at a Solution

I calculate $\displaystyle B_{ij}=\int_0^1 (1+x)[ i \pi \cos(i \pi x))(j \pi \cos(j \pi x)]=ij \pi^2 \int_0^1 (1+x) \left[\frac{1}{2} [\cos(i+j)\pi x+\cos(i-j)\pi x\right ]dx$

$\displaystyle = \frac{ij \pi^{2}}{2} \int_0^1\left [ \cos(i+j) \pi x+\cos(i-j) \pi x +x \cos(i+j) \pi x+x \cos(i-j) \pi x \right ]dx$

Now for the second and last term in the integrand if we substitute $i=j=1$ after integrating we will get a 0 in the denominator ....but the book calculates $B_{ij}=\frac{3 \pi^2}{4}$

What have I done wrong?
Did you mean to say "numerator" ?

cos(0) = 1

## Homework Statement

Folks Evaluate $B_{11}$ given

$\displaystyle B_{ij}=\int_0^1 (1+x) \frac{d \phi_i}{dx} \frac{ d\phi_j}{dx} dx$ where $\phi_i= sin i \pi x$ and $\phi_j=sin j \pi x$

## The Attempt at a Solution

I calculate $\displaystyle B_{ij}=\int_0^1 (1+x)[ i \pi \cos(i \pi x))(j \pi \cos(j \pi x)]=ij \pi^2 \int_0^1 (1+x) \left[\frac{1}{2} [\cos(i+j)\pi x+\cos(i-j)\pi x\right ]dx$

$\displaystyle = \frac{ij \pi^{2}}{2} \int_0^1\left [ \cos(i+j) \pi x+\cos(i-j) \pi x +x \cos(i+j) \pi x+x \cos(i-j) \pi x \right ]dx$

Now for the second and last term in the integrand if we substitue $i=j=1$ after integrating we will get a 0 in the denominator ....but the book calculates $B_{ij}=\frac{3 \pi^2}{4}$

What have I dont wrong?
Did you mean to say "numerator" ?

cos(0) = 1
Well no...looking at the second term $\displaystyle \int_0^1 \cos(i-j) \pi x dx= \left [\frac{sin(i-j) \pi x}{\pi(i-j)} \right]_0^1$ 1 is indeterminate....

I dont know how that answer $\frac{3 \pi^2}{4}$ was evaluated...

vela
Staff Emeritus
Homework Helper
You can't integrate it like that when i=j. Assume i=j, simplify the integrand, and then integrate.

SammyS
Staff Emeritus
Homework Helper
Gold Member
Well no...looking at the second term $\displaystyle \int_0^1 \cos(i-j) \pi x dx= \left [\frac{sin(i-j) \pi x}{\pi(i-j)} \right]_0^1$ 1 is indeterminate....

I don't know how that answer $\frac{3 \pi^2}{4}$ was evaluated...
Well, if i = j then you are integrating cos((0)πx), but cos((0)πx) = 1

You can't integrate it like that when i=j. Assume i=j, simplify the integrand, and then integrate.
Well, if i = j then you are integrating cos((0)πx), but cos((0)πx) = 1
Thank you, I thought one has to insert the index values AFTER integration....? Where is the mathematical rule that asserts this?

Cheers