SOLUTION: Make the trigonometric substitution x = a tan θ for −π/2 < θ < π/2 and a > 0. Simplify the resulting expression. (x^2/{{{sqrt (a2 + x2)}}}) I do not know how to start

Algebra ->  Trigonometry-basics -> SOLUTION: Make the trigonometric substitution x = a tan θ for −π/2 < θ < π/2 and a > 0. Simplify the resulting expression. (x^2/{{{sqrt (a2 + x2)}}}) I do not know how to start      Log On


   

Question 1159331: Make the trigonometric substitution
x = a tan θ for −π/2 < θ < π/2 and a > 0.
Simplify the resulting expression.
(x^2/sqrt+%28a2+%2B+x2%29)
I do not know how to start this problem! What am I substituting?? HELP.
Found 2 solutions by ikleyn, jim_thompson5910:
Answer by ikleyn(52824) About Me  (Show Source):
You can put this solution on YOUR website!
.

In the denominator, which is  sqrt%28a%5E2+%2B+x%5E2%29,  you should replace x by  a%2Atan%28theta%29.


It is "how to start".


Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

We are given x+=+a%2Atan%28theta%29

Wherever you see an 'x', replace it with a%2Atan%28theta%29. Then simplify.

y+=+%28x%5E2%29%2F%28sqrt%28a%5E2+%2B+x%5E2%29%29

Make the replacements

y+=+%28a%5E2%2Atan%5E2%28theta%29%29%2F%28sqrt%28a%5E2+%2B+a%5E2%2Atan%5E2%28theta%29%29%29



y+=+%28a%5E2%2Atan%5E2%28theta%29%29%2F%28sqrt%28a%5E2%281+%2B+tan%5E2%28theta%29%29%29%29 Factor out the a^2

Break up the root

y+=+%28a%5E2%2Atan%5E2%28theta%29%29%2F%28a%2Asqrt%281+%2B+tan%5E2%28theta%29%29%29 Works because a > 0

y+=+%28a%2Atan%5E2%28theta%29%29%2F%28sqrt%281+%2B+tan%5E2%28theta%29%29%29

y+=+%28a%2Atan%5E2%28theta%29%29%2F%28sqrt%28sec%5E2%28theta%29%29%29 The equation 1+tan^2 = sec^2 is one variation of the pythagorean trig identity

y+=+%28a%2Atan%5E2%28theta%29%29%2F%28sec%28theta%29%29%29 Secant is positive when theta is between -pi/2 and pi/2

y+=+a%2Atan%5E2%28theta%29%2A%28%281%29%2F%28sec%28theta%29%29%29

y+=+a%2Atan%5E2%28theta%29%2Acos%28theta%29

y+=+a%2A%28%28sin%5E2%28theta%29%29%2F%28cos%5E2%28theta%29%29%29%2Acos%28theta%29



One pair of cosine terms divide and cancel

y+=+%28a%2Asin%5E2%28theta%29%29%2F%28cos%28theta%29%29

y+=+%28a%2Asin%28theta%29%2Asin%28theta%29%29%2F%28cos%28theta%29%29

y+=+a%2Asin%28theta%29%2A%28%28sin%28theta%29%29%2F%28cos%28theta%29%29%29

y+=+a%2Asin%28theta%29%2Atan%28theta%29%29

------------------------------------------------------------

Since y+=+%28x%5E2%29%2F%28sqrt%28a%5E2+%2B+x%5E2%29%29 and y+=+a%2Asin%28theta%29%2Atan%28theta%29%29, we can say

%28x%5E2%29%2F%28sqrt%28a%5E2+%2B+x%5E2%29%29+=+a%2Asin%28theta%29%2Atan%28theta%29%29 only when a > 0 and -pi/2 < theta < pi/2.