# Trigonometric substitution

## Homework Statement

evaluate the definite integral ∫(0 to 3) dx/sqrt(25+x^2)

## The Attempt at a Solution

I first used substitution and set x=5tanθ, and dx=5tanθsecθdθ

then i wrote the integral as 5∫ tanθsecθdθ/sqrt(25(1+tan^2(θ))

after some simplification i got

∫tanθsecθdθ/secθ = ∫tanθ =tanθsecθ

I then used θ=arctan(x/5) from the original x substitution and my final solution looked like this

tan(arctan(3/5))sec(arctan(3/5))-tan(arctan(0))sec(arctan(0))

It is incorrect, can anyone let me know where i might have done something incorrectly? Thanks

## Answers and Replies

jedishrfu
Mentor
if x=5 tan(theta) then isn't dx=5sec^2(theta) dtheta

if x=5 tan(theta) then isn't dx=5sec^2(theta) dtheta

oh you're right, but after solving the integral i still end up with tanθsecθ.. Unless i am doing something wrong in finding my θ values for x=3 and x=0?

Dick
Science Advisor
Homework Helper
oh you're right, but after solving the integral i still end up with tanθsecθ.. Unless i am doing something wrong in finding my θ values for x=3 and x=0?

No, you don't get tanθsecθ. And the integral of tanθ isn't even tanθsecθ. What are you doing?

SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member

## Homework Statement

evaluate the definite integral ∫(0 to 3) dx/sqrt(25+x^2)

## The Attempt at a Solution

I first used substitution and set x=5tanθ, and dx=5tanθsecθdθ

then i wrote the integral as 5∫ tanθsecθdθ/sqrt(25(1+tan^2(θ))

after some simplification i got

∫tanθsecθdθ/secθ = ∫tanθ =tanθsecθ

I then used θ=arctan(x/5) from the original x substitution and my final solution looked like this

tan(arctan(3/5))sec(arctan(3/5))-tan(arctan(0))sec(arctan(0))

It is incorrect, can anyone let me know where i might have done something incorrectly? Thanks
If you use your substitution to find $\displaystyle \ \int \frac{dx}{\sqrt{25+x*2}}\,,\$ you get $\displaystyle \ \int\sec(\theta)\,d\theta\ .$

The anti-derivative of sec(θ) is not tan(θ)sec(θ) . The derivative of sec(θ) is tan(θ)sec(θ) . You seem to have these confused.

A better substitution uses the hyperbolic sine, x = sinh(u) .

If you use your substitution to find $\displaystyle \ \int \frac{dx}{\sqrt{25+x*2}}\,,\$ you get $\displaystyle \ \int\sec(\theta)\,d\theta\ .$

The anti-derivative of sec(θ) is not tan(θ)sec(θ) . The derivative of sec(θ) is tan(θ)sec(θ) . You seem to have these confused.

A better substitution uses the hyperbolic sine, x = sinh(u) .

oh right thanks, i used ∫secx=ln(secx+tanx) and it worked. Just a question about the hyperbolic sine, would you mind explaining it a bit to me or linking me to a good place that explains it? I don't think ive been taught it yet but one of the listed possible solutions has arccosh in it, and im curious as to what exactly it is

SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member