U-Substitution in trig integral

• ago01
In summary: This answer was wrong. The actual answer involved fully simplifying the fraction and using##u = \sin{x}##.
ago01
Homework Statement
##\int \frac{\csc{x}\cot{x}}{1+\csc^2{x}}dx##
Relevant Equations
U-Sub
##\int \frac{\csc{x}\cot{x}}{1+\csc^2{x}}dx##

Let ##u = \csc{x}##

then

##-du = \csc{x}\cot{x}dx##

So,

##\int \frac{\csc{x}\cot{x}}{1+\csc^2{x}}dx##

##-\int \frac{1}{1+u^2}du = -\arctan{u} + C##

##-\arctan{\csc{x}} + C##

This answer was wrong. The actual answer involved fully simplifying the fraction and using

##u = \sin{x}##.

I understand how this approach is done and it makes complete sense.

What I'm confused about is where I went wrong in my derivation. I don't see my mistake, and the only difference is I ended up with ##\csc{x}## and a negative sign.

Could you help me spot my error? Usually bad u-subs makes things harder so I must've made a dumb subtle mistake I can't see.

fresh_42 said:
I suspect that the mistake is hidden in the boundaries. ##\tan^{-1}(\sin(x))+\tan^{-1}(\csc(x))## is a step function:
https://www.wolframalpha.com/input?i=arctan(sin(x))+++arctan(1/sin(x))=

Interesting...I suppose this makes it clear why you should simplify as much as possible to get to the meat of the integral. The graphs are only equal at the "peaks". Fascinating.

ago01 said:
Interesting...I suppose this makes it clear why you should simplify as much as possible to get to the meat of the integral. The graphs are only equal at the "peaks". Fascinating.
What peaks?

One issue here is that the constants of integration differ. Another issue is that the integrand for the integral
##\displaystyle \int \frac{\csc{x}\cot{x}}{1+\csc^2{x}}dx ##
is undefined at ## x = k\pi##, where ##k## is an integer.. On the other hand the integrand for
##\displaystyle \int \frac{\cos{x}}{1+\sin^2{x}}dx ##
it is defined for all ##x## .

The graph suggested by @fresh_42 shows that
##\displaystyle \tan^{-1}(\sin(x))##
and
##\displaystyle -\tan^{-1}(\csc(x)) ##
differ by
##\pm \dfrac \pi 2 ## .

Last edited:
ago01

What is U-Substitution in trigonometric integrals?

U-Substitution is a method used to solve integrals involving trigonometric functions. It involves substituting a variable, often denoted as u, for a portion of the integral in order to simplify the problem and make it easier to solve.

When should I use U-Substitution in trigonometric integrals?

U-Substitution is most useful when the integral contains a product of a trigonometric function and its derivative, or when the integral contains a complicated trigonometric expression that can be simplified using a substitution.

How do I perform U-Substitution in trigonometric integrals?

To perform U-Substitution, you first need to identify the portion of the integral that can be substituted with a variable. Then, you substitute the variable and its derivative into the integral, making sure to adjust the limits of integration if necessary. Finally, you solve the integral with the substituted variable and then replace it with the original expression.

What are some common mistakes to avoid when using U-Substitution in trigonometric integrals?

One common mistake is forgetting to adjust the limits of integration when substituting a variable. It is also important to choose an appropriate variable substitution and to correctly evaluate the integral with the substituted variable. Additionally, it is important to check the final answer by differentiating it to ensure it is correct.

Are there any other methods for solving trigonometric integrals besides U-Substitution?

Yes, there are other methods such as integration by parts and trigonometric identities. It is important to understand when to use each method and to practice solving a variety of integrals to become familiar with their differences and similarities.

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