7–84. Evaluate the following integrals.
13. ∫ [1 / (eˣ √...

Question

7–84. Evaluate the following integrals.

13. ∫ [1 / (eˣ √(1 – e²ˣ))] dx

Accepted Answer

Hello. In this video, we are going to be evaluating the given integral. We are given the general integral of 1 divided by the quantity, 2 raised to the power of X, multiplied by the square root of 1 minus 4 raised to the power of XD X. Now, in order to approach this type of integral, let's go ahead and approach this problem by using a U substitution. We will allow you to equal to 2 power of X. And if we were to find a substitution for DX in terms of you, we will end up with DU is equal to the natural logarithm of 2. Multiplied by 2 power of XDX. Now, we need to find a substitution of DX into terms of U. Let's go ahead and simplify this problem by moving LN 2 and to the power of X to the left hand side of the statement. This will leave us with D X is equal to 1 divided by 2 power of X multiplied by the natural logarithm of 2Du. Now again, we did say that U is equal to 2 power of X. So we can resubstitute 2 power of X with U, and that would give us DX is equal to 1 divided by U multiplied by the natural logarithm of 2 D U. Now, this gives us a substitution for DX and we are going to substitute 2 to the power of X into terms of U. But what about this 4 to the power of X? While recall that 4 to the power of X can be rewritten as 2 squad to the power of X, and furthermore that can be rewritten or reordered as 2 to the power of X squared. That means that since we know that U is equal to 2 to the power of X, we can rewrite 4 of X as U squad. So this is going to be the substitution for 4 power of X. So now if we were to apply all of our substitutions, we will end up with the integral of 1 divided by U multiplied by the square root of 1 minus U squared multiplied by 1 divided by U multiplied by the natural logarithm of 2 D U. Now, if we were to move 1 divided by the natural logarithm of 2 to the front of the integral as a constant, and multiply the U terms together, we can further simplify this integral to be 1 divided by the natural logarithm of 2. Multiplied by the integral of 1 divided by U2, multiply by the square root of 1 minus U 2 DU. Now, this looks like a complicated problem, but notice that we can approach this integral by using trigonometric substitution. This is in the form of a sign trigonometric substitution. So if we were to use trigonometric substitution, we can pick a separate variable. Let's say, for example, V. And we will let V equal to sine of theta. Now, we need to solve DU in terms of V. So by taking the derivative of V with respect to theta, we get DV is equal to cosine of theta D theta. Now, if we were to apply our trigonometric substitution, we can rewrite the integral to be 1 divided by the natural logarithm of 2, multiplied by the integral of 1 divided. By sine theta squared or sine squared theta. Multiplied by the square root of 1 minus. Sin square theta. And this will be multiplied by the substitution of DU into terms of V, so this will go ahead and be cosine theta. The theater This is going to be the full trigonometric substitution, and now what we need to do is we need to go ahead and simplify. First, by using trigonometric identities, we can rewrite 1 minus sine square theta to be cosine squared theta, and cosine square theta underneath the square root will leave us with just cosine. So we have 1 divided by the natural logarithm of 2, multiplied by the integral of 1 divided by sin squared theta. Cosine theta, and this is still going to be multiplied by cosine theta D theta. If we multiply the cosine to the numerator, then we have cosine theta in the numerator and D theta. Notice that we have cosine in both the numerator and denominator. We can reduce this to just be 1, leaving us with 1 divided by the natural logarithm of 2 multiplied by the integral of 1 divided by sine squared theta d theta. One divided by sin squared theta can be rewritten as cosequent squared theta. So, We, by using trigonometric identities again, we have 1 divided by the natural logarithm of 2 multiplied by the integral of cosequent squared theta d theta, and cosequent squared theta is the antiderivative or is the derivative of tangent. So the cotangent apologies. So that means that the anti-derivative, this is going to be negative cotangent. So we have negative, 1 divided by the natural logarithm of 2 cotangent of theta plus C. Now what we need to do is we need to rewrite cotangent data back in terms of V, then back in terms of you. Now originally, we said that V is equal to sin of theta. If we were to rewrite this into a right triangle. With respect to sin theta. This is going to give us a right triangle where the opposite side is going to be V, and the hypotenuse is 1, and by using the Pythagorean theorem, we can solve the missing side to be the square root of 1 minus V squared. And cotangent of this right triangle is going to be the square root of 1 minus V squared. Divided by V. So, we can further rewrite this integral to be negative 1 divided by the natural logarithm of 2, multiplied by the quantity, the square root of 1 minus V2 divided by V plus C. And if we were to back substitute V with our original U substitution, which was 2 to the power of X, then we can rewrite the integral to be negative 1 divided by the natural logarithm of 2 multiplied by the square root of 1 minus 4 power of X divided by 2 power of X plus C. And multiplying both of these values together is going to give us the negative square root of 1 minus 4 to the power of x divided by 2 power of X multiplied by the natural logarithm of 2 plus C. and this is going to be the simplest form of the solution to the indefinite integral. And with that being said, the overall solution is going to be B. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

7–84. Evaluate the following integrals.
13. ∫ [1 / (eˣ √(1 – e²ˣ))] dx

Key Concepts

Integration

Substitution Method

Exponential Functions

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