Make the trigonometric substitution x = a tan θ for −π/2 < θ < π/2 and a > 0

Question

Simplify the resulting expression: x^2/sqrt(a^2+x^2).

 

I actually have no idea how to go about solving this at all. If anyone could be kind enough to explain this to me step by step, I'll be really grateful! I just don't totally understand what this question means.

Answer 1

Start with the expression

x²/sqrt(a²+x²)

 

Then, replace x with a*tan(θ)

 

(a*tan(θ))²/sqrt(a²+(a*tan(θ))²)

 

Now, see how much you can simplify the expression.

 

a²*tan²(θ)/sqrt(a²+a²*tan²(θ))

*Note that the convention is to write (tan(θ))² as tan²(θ)

 

a²*tan²(θ)/sqrt(a²*(1+tan²(θ))

 

Then use the trig identity: 1+tan²(θ)=sec²(θ)

And take the square root

*Here's where the a>0 comes in--since a>0, sqrt(a^2)=a, not -a

*Since -π/2<θ<π/2, sec(θ)>0, and sqrt(sec²(θ))=sec(θ)

 

a²*tan²(θ) / a*sec(θ)

 

Then, cancel out an a

and note that 1/sec(θ)=cos(θ)

 

a*tan²(θ)*cos(θ)

 

Now, since tan(θ)=sin(θ)/cos(θ), we can write

 

a*tan(θ)*sin(θ)

Answer 2

x=atan

simplify x^2/√(a^2+x^2)

replace x by atan

(atan)^2/√(a^2+[atan]^2)

a^2tan^2/√(a^2+a^2tan^2)

a^2tan^2/√a^2(1+tan^2)

a^2tan^2/a√(1+tan^2)

atan^2/√(1+tan^2)

atan^2/√sec^2

atan^2/sec

(asin^2)/(cos^2)÷sec

(asin^2)/(cos^2)÷(1/cos)

(asin^2)/(cos^2)*(cos)

(asin^2)/(cos)

(a)(sin)(sin)/(cos)

(a)(sin)(sin/cos)

asinθtanθ