3. Evaluate the integral| Vz? +edr,where c can be negative or positive, without trigonometric substitutions.(a) If c > 0 what trig substitution you would use and why?(b) If c < 0 what trig substitution you would use and why?(c) (Solution #1) Use the integration by parts with u = vx² + c and reduce this integral tothe one found in Problem 2. Evaluate the integral.(d) (Solution #2) Make the substitutionVx² + c =t – r.Hint: Square both sides and simplify. Find r in terms of t. Find vr² + c = t – x = ....Find dæ. After plugging everything into the integral just divide numerator by denominatorto obtain simple integrals.(e) Evaluate the integral and return to variable r.(f) Did you get the same solution as in part (c)?Remark: The substitution suggested in Problems 2 and 3 is a particular case of the so-calledEuler's substitutions. You can find more on them online.

Answered: | V² +cdr,

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3. Evaluate the integral
| Vz? +edr,
where c can be negative or positive, without trigonometric substitutions.
(a) If c > 0 what trig substitution you would use and why?
(b) If c < 0 what trig substitution you would use and why?
(c) (Solution #1) Use the integration by parts with u = vx² + c and reduce this integral to
the one found in Problem 2. Evaluate the integral.
(d) (Solution #2) Make the substitution
Vx² + c =t – r.
Hint: Square both sides and simplify. Find r in terms of t. Find vr² + c = t – x = ....
Find dæ. After plugging everything into the integral just divide numerator by denominator
to obtain simple integrals.
(e) Evaluate the integral and return to variable r.
(f) Did you get the same solution as in part (c)?
Remark: The substitution suggested in Problems 2 and 3 is a particular case of the so-called
Euler's substitutions. You can find more on them online.

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(As per bartleby guidelines, for questions containing multiple subparts, only first three subparts will be solved. Please upload other parts separately.)

Consider the given integral,

$I = \int \sqrt{x^{2} + c} d x$

Concept: The purpose of using a trigonometric substitution is to simplify the function before integration. Here the square root function may cause a tedious calculation. So the goal is to get rid of the square root function.

For this, it is required that the expression inside the root be written as a square of something. Here comes the need of a proper substitution.

By using trigonometric substitution, it is aimed to take advantage of the following identities,

$\sin^{2} (θ) + \cos^{2} (θ) = 1 \dots \dots \dots (i) \tan^{2} (θ) + 1 = \sec^{2} (θ) \dots \dots \dots (i i) 1 + \cot^{2} (θ) = \csc^{2} (θ) \dots \dots \dots (i i i)$

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