When integrating by trig substitution why do you use what you use?

[brandy]

when integrating by trig substitution why do you use what you use??

for example int. (1+x^2)^0.5 dx

why do you use x= tan u
i mean obviously because it works, but if you didn't know it works how would you figure it out?

i would think that you should use x=sinh u
but I've been trying that and its not what wolfram alpha gets.

 

[Some Pig]

Both works,
Let $$x=\sinh u,\ dx=\cosh udu$$
$$\int\sqrt{1+x^2}\ dx$$
$$=\int\cosh^2udu$$
$$=\int\tfrac12(1+\cosh2u)du$$
$$=\tfrac12u+\tfrac14\sinh2u+c$$
$$=\tfrac12\sinh^{-1}x+\tfrac12x\sqrt{1+x^2}+c$$
Exactly what Wolfram got.

 

[brandy]

oh. haha. i just can't do maths is all.
haha,
thanks anyway.

 

[brandy]

sorry, how did you sub u into sinh(2u)
isnt that just sinh(2sinh^-1(x)?
how did you simplify?

 

[HallsofIvy]

First, since this has nothing to do with differential equations, I am moving this thread to "Calculus and Analysis".

Second, the idea for all the trig functions is that
[tex]sin^2(u)+ cos^2(u)= 1[/itex]
and its variations:
dividing through by $cos^2(u)$
$$tan^(u)+ 1= sec^2(u)$$
and dividing through instead by $sin^2(u)$
$$1+ cot^2(u)= csc^2(u)$$

Since $sin^(u)+ cos^2(u)= 1$ can also be written as $cos^2(u)= 1- sin^2(u)$, replacing the x is $1- x^2$ sin(u) gives a "square", $cos^2(u)$ allowing us to get rid of the square root:
$$\int\sqrt{1- x^2} dx= \int \sqrt(1- sin^2(u))(cos(u) du)= \int\sqrt{cos^2(u)}(cos(u)du)= \int cos^2(u)du$$

Similarly because [itex]sec^2(u)= 1+ tan^2(u)[/tex] any "$1+ x^2$", with [itex]x= tan(u)[/tex], becomes $sec^2(u)$.


The hyperbolic functions also work because
$$cosh^2(u)- sinh^2(u)= 1$$
etc.

 

[brandy]

thanks guys!

 

[Mark44]

Trig substitutions are very helpful when the integrand contains a sum or difference of squares: x² + a², x² - a², or a² - x², and especially when the integrand contains the square root of one of these three. I'll focus on integrands with the square root of one of these three expressions, but the idea is more general than that.

Rather than memorize which trig substitution goes with which form, I draw a right triangle and label the sides and hypotenuse in accordance with the expression I'm dealing with. I label the acute angle θ.


$$\sqrt{x^2 + a^2}$$

This expression represents the hypotenuse of the right triangle. You can label the two other sides as x and a in either combination, but most texts label the altitude as x and the base as a. This gives tan(θ) = x/a, or a tan(θ) = x. From this substitution you can get the relationships between the differentials: dx = a sec²(θ)dθ and another relationship that involves the radical; namely, sec(θ) = sqrt(x² + a²)/a, or a*sec(θ) = sqrt(x² + a²).

$$\sqrt{x^2 - a^2}$$
Here the radical suggests that it represents one of the sides of the triangle, with x being the length of the hypotenuse and sqrt(x² - a²) being one of the sides. Many texts pick the altitude for this value and label the base as a.

This gives cos(θ) = a/x, or equivalently, sec(θ) = x/a, or a sec(θ) = x. From this you get dx = a sec(θ) tan(θ) dθ. An expression involving the radical is a*tan(θ) = sqrt(x² - a²).

$$\sqrt{a^2 - x^2}$$
This is similar the one above, but the hypotenuse is labelled a. The two sides can be labelled as x and sqrt(a² - x²) in either way, but most often I've seen it with the altitude labelled as x and the base labelled with the radical.

This gives sin(θ) = x/a, or a sin(θ) = x, so dx = a cos(θ) dθ. An expression for the radical is cos(θ) = sqrt(a² - x²)/a, so a*cos(θ) = sqrt(a² - x²)

 

[paulfr]

Mark44 very nice explanation

I have all these combinations of quadratic binomials worked out
in detail on the 2nd page of Calculus Tables041111 at
www.scribd.com/pfreda