Calculus Trig Sub INTEGRALS HELP!!! Please!
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Calculus Trig Sub INTEGRALS HELP!!! Please!

[From: ] [author: ] [Date: 11-09-21] [Hit: ]
......

2tanθ secθ - 2 ln |tanθ + secθ| + C

let's now convert this back to x variable, recalling that tanθ = x/2;

hence:

secθ = √(1 + tan²θ) = √[1 + (x/2)²] = √[1 + (x²/4)] = √[(4 + x²) /4] = [√(4 + x²)] /2

then, substituting back:

2tanθ secθ - 2 ln |tanθ + secθ| + C = 2(x/2) {[√(4 + x²)] /2} - 2 ln |(x/2) +
{[√(4 + x²)] /2}| + C =

(x/2)√(4 + x²) - 2 ln |[x + √(4 + x²)] /2| + C =

(applying logarithm properties)

(x/2)√(4 + x²) - 2 [ln |x + √(4 + x²)| - ln (2)] + C =

(x/2)√(4 + x²) - 2 ln |x + √(4 + x²)| + 2 ln (2) + C

being 2 ln (2) a mere constant, we can include it in C, ending with:


∫ [x² /√(4 + x²)] dx = (x/2)√(4 + x²) - 2 ln [x + √(4 + x²)] + C

(the absolute value is no needed any longer since the argument x + √(4 + x²) is always positive)


I hope it's helpful
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