how does the integral of cscx = ln|cscx - cotx|+C and ln |tan x/2|+C?
Well, it seems like you're getting yourself into quite a tangle with trigonometric integrals! Let me help unravel the mystery for you.
First of all, the integral of csc(x) is indeed a bit tricky. One way to approach it is by using a nifty trigonometric identity: csc(x) is the same as 1/sin(x). So let's rewrite the integral as ∫(1/sin(x)) dx.
Now, to tackle this, we can use a clever substitution. Let u = cos(x), du = -sin(x) dx. We can then convert the integral into ∫(-1/(u*(u^2-1))) du.
Now, this can be rewritten as ∫(-1/(u^3-u)) du. Impressed by my mathematical prowess, yet?
Next, we can use partial fraction decomposition to simplify it further. By breaking down the integrand, we get ∫(1/(u+1) - 1/u - 1/(u-1)) du.
Time for integration! This gives us the following expression: ln|u+1| - ln|u| - ln|u-1| + C.
Finally, we can substitute back to get the original variable x, which gives us ln|cos(x) + 1| - ln|cos(x)| - ln|cos(x) - 1| + C.
But wait, there's more! Using a nifty logarithmic property, we can simplify this expression further. ln(a) - ln(b) is the same as ln(a/b). Applying this to our integral, we have ln|(cos(x) + 1)/(cos(x))| - ln|((cos(x) - 1))/(cos(x))| + C.
Now, this can be simplified even further. Remember another cool trigonometric identity: tan(x/2) = (1-cos(x))/(sin(x)). So the expression becomes ln|tan(x/2)| + C.
Voila! We made it. The integral of csc(x) is ln|csc(x) - cot(x)| + C, and also ln|tan(x/2)| + C. Keep up the great work, and remember, math is an adventure!
To understand how the integral of csc(x) can be expressed as ln|csc(x) - cot(x)| + C and ln|tan(x/2)| + C, we need to follow these steps:
Step 1: Rewrite csc(x) as 1/sin(x).
Step 2: Use the substitution method, by letting u = sin(x). Then, du = cos(x) dx.
Step 3: Rewrite the integral using the substitution from step 2 as ∫(1/u)(du/cos(x)).
Step 4: Simplify the expression to ∫(du/u)(1/cos(x)).
Step 5: Since 1/cos(x) is the same as sec(x), rewrite the expression as ∫(du/u)(sec(x)).
Step 6: Now, rewrite sec(x) as 1/cos(x), so we have ∫(du/u)(1/cos(x)/cos(x)).
Step 7: Cancel out the cos(x) terms to get ∫du/u.
Step 8: The integral of du/u is ln|u| + C.
Step 9: Replace u with sin(x) again to obtain ln|sin(x)| + C as the result.
Therefore, the integral of csc(x) is ln|sin(x)| + C.
Now, let's see how it can also be expressed as ln|csc(x) - cot(x)| + C and ln|tan(x/2)| + C.
Step 10: Start with the expression ln|sin(x)| + C and rewrite it as ln|cos(x)/sin(x)| + C.
Step 11: Use the quotient rule of logarithms, which states that ln(a/b) = ln(a) - ln(b), to rewrite the expression as ln|cos(x)| - ln|sin(x)| + C.
Step 12: Recall the trigonometric identity cot(x) = cos(x)/sin(x), and rewrite the expression as ln|cot(x)| - ln|sin(x)| + C.
Step 13: Use the logarithmic identity ln(a) - ln(b) = ln(a/b), to combine the logarithmic terms as ln|cot(x)/sin(x)| + C.
Step 14: Rewrite cot(x)/sin(x) using the reciprocal identity, which states that cot(x) = 1/tan(x), so cot(x)/sin(x) is equal to 1/(sin(x)/cos(x)), which is cos(x)/sin(x).
Step 15: Substitute cos(x)/sin(x) with csc(x), so we have ln|csc(x)| + C as the result.
Therefore, the integral of csc(x) can also be expressed as ln|csc(x)| + C, which is equivalent to ln|csc(x) - cot(x)| + C and ln|tan(x/2)| + C.
To understand why the integral of csc(x) is ln|csc(x) - cot(x)| + C, let's go through the steps to solve the integral. We'll also see the connection to ln|tan(x/2)| + C at the end.
Step 1: Rewrite csc(x) using trigonometric identities.
Recall that csc(x) is the reciprocal of sin(x). We can rewrite csc(x) as 1/sin(x).
Step 2: Start by multiplying and dividing by (csc(x) + cot(x)).
To simplify the integral, multiply and divide by (csc(x) + cot(x)). This is a common technique called multiplying by the conjugate.
∫ csc(x) dx = ∫ (1/sin(x)) [(csc(x) + cot(x))/(csc(x) + cot(x))] dx
Step 3: Simplify the expression.
Using the multiplication property of fractions, we can simplify the expression in the integrand:
∫ (csc(x) + cot(x))/(sin(x)(csc(x) + cot(x))) dx
Since sin(x)(csc(x) + cot(x)) cancels out, we have:
∫ dx = ∫ 1 dx
Step 4: Integrate the simplified expression.
The integral of 1 is simply x:
∫ dx = x + C
So, we have:
∫ csc(x) dx = ln|csc(x) + cot(x)| + C
But wait! We obtained a different expression than you asked about. However, the expression ln|csc(x) + cot(x)| is equivalent to ln|csc(x) - cot(x)| because cot(x) is the reciprocal of tan(x), and we can express cot(x) as -cot(x). So:
∫ csc(x) dx = ln|csc(x) - cot(x)| + C
Now, let's discuss the connection to ln|tan(x/2)| + C.
By using the half-angle identity of tangent, we can rewrite csc(x) - cot(x) as 2/tan(x/2).
Therefore, ln|csc(x) - cot(x)| + C can be written as ln|2/tan(x/2)| + C. Simplifying further, we get:
ln|2/tan(x/2)| + C = ln|2| - ln|tan(x/2)| + C = ln|tan(x/2)| + ln|2| + C.
Since ln|2| + C is a constant, we can write the final result as:
∫ csc(x) dx = ln|tan(x/2)| + C.
So, both ln|csc(x) - cot(x)| + C and ln|tan(x/2)| + C are correct expressions for the integral of csc(x), just written in different forms.