Simplifying Integrals with Trigonometric Substitutions (Part II)

Trigonometric Substitutions (Part II)

The method of substitution by trigonometric function can be applied to another type of integral. Here is an example of such an integral.

Example: ∫dxb2x2−c2, x >cb

In order for the expression inside the square root sign to be non-negative, we must have x greater than c over b. We can simplify this expression by taking out B and therefore it is equal to b times x squared minus c over b squared.

b2x2 − c2 = b x2 − (cb)2 = b x2 − a2 with a =cb

In order for the expression inside the square root sign to be nonnegative, we must have x greater than c over b. Therefore, we can simplify a little bit by taking out B and therefore it is equal to b times x squared minus c over b squared.

x = a sec θ, 0 < θ <π2and dx = a sec θ tan θ dθ

Now, in a situation similar to those described above, we seek to find a trigonometric substitution that will allow us to rewrite x2 − a2 as a square.

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Now, since it is clear that x must be equal to some value of secant theta , we choose x = a times secant theta with its range being between 0 and 2π . We will see why this choice is appropriate shortly.

x2 − a2 = a2(sec2θ − 1) = a2tan2θ = a|tan θ | = a tan θ

So we use the secant function on both sides of the equation.

This allows us to make a substitution so that the equation becomes a perfect square, which we can then solve for its root. However, there is an absolute value sign with the tangent function in this range; we must therefore find a range for θ where tanθ ≠ 0 so that we can remove the absolute value signs from both sides of the equation.

∫a sec θ tan θ dθba tan θ =1b∫ sec θ dθ

The original integral can be transformed into this new one, where the bottom is b times a tangent of theta. What is dx? The choice of x is here, so you differentiate both sides and get dx equals a differentiate second theta. After some cancellation you end up with this integral which we have already seen before that it equals here.

∫a sec θ tan θ dθba tan θ =1b∫ sec θ dθ

=1bln| sec θ + tan θ | + C

Finally, we have to find the length of the hypotenuse by applying Pythagoras's theorem.

Transform the terms in this equation back to x.

∫a sec θ tan θ dθba tan θ =1b∫ sec θ dθ

=1bln|sec θ + tan θ | + C

=1bln xa+x2−a2a||||||+ C

The integral of secant theta x over a, times tangent theta, can be found using trigonometric substitutions. This is done by first finding the substitution for x in terms of a. Then use this substitution with tangent theta and secant theta to find the integral.

Substituting with x = a tan θ

a2+ x2 = a2+ a2tan2θ = a21 + tan2θ( ) = a2sec2θ

a2+ x2 = a|sec θ |

The first is that x equals the tangent of θ and it allows us to rewrite this expression as a squared plus x squared to become a square.

with x = a sin θ

− x2 = a2 − a2sin2θ = a21 − sin2( θ) = a2cos2θ

a2 − x2 = a|cos θ |

To express a square minus x, we substitute x with a sine theta and arrive at the expression above.

with x = a sec θ

x2 − a2 = a2sec3θ − a2 = a2sec2( θ − 1) = a2tan2θ

x2 − a2 = a|tan θ |

To express the function of this shape, we use x equals a secant times theta. And this makes x squared minus a squared equals a tangent squared times theta squared. So that we can take the square root immediately and convert back to functions in terms of x. In one case we make use of such a right angle triangle with two sides equal to x and a. For the second case we use this right angle triangle with two sides equal to x and a.

For the last case, we use this right-angled triangle in which x is the hypotenuse, a is the horizontal side and we can convert back things to functions of x.