How to Solve Integrals for Beginners

[Physicsisfun2005]

integral of 1/(cscx-1)


i can't think of any trig subs or anything...natural log prob? ;)...i udnno I'm truly clueless. all i need is a hint or more if u like ;)








also, am i using integrals for this prob:

Compute the average value of f(x)=x/(x+3) over the interval [-a,a], 0<a<3

I think i will just get an expression for the answer

 

[Benny]

Here is what I have done. So you should go through my working to check for errors.

<br /> \int {\frac{1}{{\cos ecx - 1}}} dx<br />

<br /> = \int {\frac{1}{{\left( {\frac{1}{{\sin x}} - 1} \right)}}} dx<br />

<br /> = \int {\frac{1}{{\left( {\frac{{1 - \sin x}}{{\sin x}}} \right)}}} dx<br />

<br /> = \int {\frac{{\sin x}}{{1 - \sin x}}dx} <br />

When I see something like the above I just think, something needs to be done to the denominator. I would use a technique similar to rationalising a surd expression as follows.

<br /> = \int {\frac{{\sin x}}{{1 - \sin x}} \times \frac{{1 + \sin x}}{{1 + \sin x}}dx} <br />

<br /> = \int {\frac{{\sin x + \sin ^2 x}}{{1 - \sin ^2 x}}dx} <br />

Now use some trig identities as follows.

<br /> = \int {\frac{{\sin x + \frac{1}{2} - \frac{1}{2}\cos 2x}}{{\cos ^2 x}}} dx<br />

<br /> = \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \frac{1}{2}\int {\frac{1}{{\cos ^2 x}}dx} } - \frac{1}{2}\int {\frac{{\cos 2x}}{{\cos ^2 x}}dx} <br />

<br /> = \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \frac{1}{2}\int {\frac{1}{{\cos ^2 x}}dx} } - \frac{1}{2}\int {\frac{{2\cos ^2 x - 1}}{{\cos ^2 x}}} dx<br />

<br /> = \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \frac{1}{2}\int {\frac{1}{{\cos ^2 x}}dx} } - \frac{1}{2}\int {\frac{{2\cos ^2 x}}{{\cos ^2 x}}} dx - \frac{1}{2}\int {\frac{{\left( { - 1} \right)}}{{\cos ^2 x}}dx} <br />

<br /> = \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \frac{1}{2}\int {\frac{1}{{\cos ^2 x}}dx} } - \int {dx + \frac{1}{2}\int {\frac{1}{{\cos ^2 x}}dx} } <br />

<br /> = \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \int {\frac{1}{{\cos ^2 x}}} - \int {dx} } <br />

<br /> = \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \int {\sec ^2 x} dx - \int {dx} } <br />

The first integral in the above line can be done by substitution or simply by inspection which can save time.

<br /> = \frac{1}{{\cos x}} + \tan x - x + c<br />

<br /> = \sec x + \tan x - x + c<br />

 

[dextercioby]

Yes,Benny,good work... Though you could have shortcutted the part with the double angle on the cosine...

Daniel.

 

[Benny]

Thanks dextercioby.

I see what you mean, I went around in circles. The question could have been finished as follows.

<br /> \int {\frac{{\sin x + \sin ^2 x}}{{1 - \sin ^2 x}}} dx<br />

<br /> = \int {\frac{{\sin x + \sin ^2 x}}{{\cos ^2 x}}} dx<br />

<br /> = \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \int {\tan ^2 x} dx} <br />

<br /> = \int {\frac{{\sin x}}{{\cos ^2 x}}dx + \int {\left( {\sec ^2 x - 1} \right)} dx} <br /> The integral on the left can be integrated by substitution or using the fact that \frac{d}{{dx}}\left( {\sec x} \right) = \tan x\sec x

So \int {\frac{1}{{\cos ecx - 1}}dx} = \sec x + \tan x - x + c.

 

[dextercioby]

I was thinking about something else...Right from the beginning...It doesn't really matter.It would have been another (more simple) trick besides the one you pulled...

Daniel.

P.S.These 2 tricks shorten the calculus,but the disadvantage is that u may not see either of them...

 

[poolwin2001]

In the numerator you could have used Sinx=-(-Sinx)=-(1-Sinx-1)
which would have simplified the calc instead of *ing and /ing by 1+Sinx

 

[dextercioby]

In the numerator you could have used Sinx=-(-Sinx)=-(1-Sinx-1)

OMG,how can u say that?? Would u agree that what u said is basically an alambicated way of:
\sin x=(\sin x-1)+1

and the minus would appear naturally (when symplifying the denominator and the numerator),not forced...??

Daniel;

PS.Benny,that's the trick i mentioned...

 

[poolwin2001]

What do you mean by natural and forced??
Yes I was talking about this only.\sin x=(\sin x-1)+1
I added the -ve sign to prevent confusion,but it seems that it didn't work.