87-92. An integrand with trigonometric functions in the numera...

Question

87-92. An integrand with trigonometric functions in the numerator and denominator can often be converted to a rational function using the substitution u = tan(x/2) or, equivalently, x = 2 tan⁻¹u. The following relations are used in making this change of variables.
A: dx = 2/(1 + u²) du
B: sin x = 2u/(1 + u²)
C: cos x = (1 - u²)/(1 + u²)
88. Evaluate ∫ dx/(2 + cos x).

A: dx = 2/(1 + u²) du

B: sin x = 2u/(1 + u²)

C: cos x = (1 - u²)/(1 + u²)

88. Evaluate ∫ dx/(2 + cos x).

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to evaluate the integral of DX divided by the quantity of 2 plus cosine of X in quantity, using partial fractions. So the first step is to actually make a substitution. So we'll set U equal to the tangent of the quantity of X divided by 2 in quantity, and making the substitution, we know that DX is going to be equal to 2 divided by the quantity of 1 plus U squad in quantity DU. And we can also rewrite cosine of X as being equal to the quantity of 1 minus U2 in quantity, divided by the quantity of 1 plus U2 in quantity. So that means that our integral of D X divided by the quantity of 2 plus cosine of X in quantity is going to be equal to the integral of 1 divided by the quantity of 2 plus the quantity of 1 minus u squad in quantity divided by the quantity of 1 plus U squad in quantity, and this gets multiplied by 2 divided by the quantity of 1 plus u squared and quantity d u. So now we just need to simplify this expression. In doing so, we see that this is going to be equal to the integral of 2 divided by the quantity of 3 plus U squared in quantity DU. So now we want to try to use partial fractions in order to simplify the integram. And so that means that 2, divided by the quantity of 3 plus U squared. And quantity, it's going to be equal to. A multiplied by U plus B, and that quantity is going to be divided by the quantity of 3 plus u squared in quantity. And we can't really do any better than this, since we can't really reduce the polynomial and the denominator any further. So based on this expression, we see that A is going to be equal to 0, and B is going to be equal to 2. So we really cannot use partial fractions to reduce our integram into two separate fractions. But what we can do is factor the 2 out in front of the integral. In doing so, this is going to be equal to 2 multiplied by the integral of 1 divided by the quantity, and here we'll rewrite the denominator a little differently. We'll have the quantity of the square root of 3 and quantity squared, plus u squared and quantity dU. Now recall that the integral of 1 divided by the quantity of a 2 plus U2 in quantity D U is equal to 1 divided by a multiplied by the arc tangent. Of the quantity of you divided by A in quantity plus C. So, in our case, A is going to be equal to the square root of 3. So that means our integral is going to be equal to. 2 multiplied by 1 divided by the square root of 3. Multiplied by the arctangent. Of the quantity of you divided by the square root of 3 in quantity plus C. So now what we want to do is to substitute back in our original expression for you in terms of X. So doing so, this is going to be equal to 2, divided by the square root of 3, multiplied by the arc tangent. Of the quantity. Of the tangent of the quantity of X divided by 2, in quantity, divided by the square root of 3, in quantity plus C. So, next we want to rewrite 1 divided by the square root of 3, as the square root 3 divided by 3. In doing so, this is going to be equal to. 2 multiplied by the square root of 3, divided by 3, multiplied by the arc tangent. Of the quantity of the square root of 3 multiplied by the tangent of the quantity of X divided by 2 in quantity divided by 3 in quantity plus C. So this here is the answer that we're looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

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