integral from 0 to pi/4 of (secxtanx dx)
Please show how to find the antidervative of secx and tanx.
Well the derivative of sec(x) = sec(x) tan(x) dx, as it's one of the formulas usually memorized. So the integral of sec(x) tan(x) dx = sec(x).
But if you want a longer method, we let sec(x) = 1/cos(x) and tan(x) = sin(x) / cos(x). Substituting,
∫ sec(x) tan(x) dx
∫ (1/cos(x)) (sin(x)/cos(x)) dx
∫ (sin(x) / cos^2 (x)) dx
Let u = cos(x)
Thus du = -sin(x) dx.
∫ -du / u^2
= 1 / u
= 1 / cos(x)
= sec(x)
It's evaluated from 0 to π/4. Thus,
= sec(π/4) - sec(0)
= sqrt(2) - 1
hope this helps~ `u`
thank you!
To find the antiderivative of sec(x), we can use a technique called integration by substitution. Here's how you can do it step by step:
Step 1: Let u = sec(x) + tan(x)
Differentiate both sides with respect to x:
du/dx = (sec(x)tan(x) + sec^2(x))
Step 2: Rearrange the equation to solve for sec(x)tan(x):
sec(x)tan(x) = du/dx - sec^2(x)
Step 3: Substitute sec(x)tan(x) in the original integral:
∫ sec(x)tan(x) dx = ∫ (du/dx - sec^2(x)) dx
Step 4: Cancel out the dx terms:
∫ sec(x)tan(x) dx = ∫ du - ∫ sec^2(x) dx
Step 5: Integrate each term separately:
∫ sec(x)tan(x) dx = u - ∫ sec^2(x) dx
Step 6: For the second term, we can use a trigonometric identity: sec^2(x) = 1 + tan^2(x)
∫ sec(x)tan(x) dx = u - ∫ (1 + tan^2(x)) dx
Step 7: Simplify the integral:
∫ sec(x)tan(x) dx = u - ∫ 1 dx - ∫ tan^2(x) dx
∫ sec(x)tan(x) dx = u - x - ∫ tan^2(x) dx
Step 8: Recall that the integral of tan^2(x) can be found using the power rule:
∫ tan^2(x) dx = (1/3)tan^3(x)
Step 9: Substitute back u = sec(x) + tan(x):
∫ sec(x)tan(x) dx = u - x - (1/3)tan^3(x)
And that's how you find the antiderivative of sec(x)tan(x).
To find the antiderivative of the function sec(x)tan(x), we can use the integral formula for the product of two functions: ∫(u*v)dx = u∫vdx + v∫udx.
Let's start by finding the antiderivative of sec(x). We can use the formula: ∫sec(x)dx = ln|sec(x) + tan(x)| + C.
Next, we'll find the antiderivative of tan(x). We can rewrite tan(x) as sin(x)/cos(x) and use the substitution method.
Let u = cos(x), then du = -sin(x)dx, and dx = -du/sin(x).
Substituting these values in, we have:
∫tan(x)dx = ∫(sin(x)/cos(x))dx = -∫(1/u)du = -ln|u| + C = -ln|cos(x)| + C.
Now that we have the antiderivatives, let's compute the integral of sec(x)tan(x):
∫(sec(x)tan(x))dx = ∫sec(x)dx * tan(x) + ∫tan(x)dx * sec(x)
Using the antiderivatives we found, we have:
= [ln|sec(x) + tan(x)|] * tan(x) + [-ln|cos(x)|] * sec(x) + C
Finally, substitute the limits of integration (from 0 to π/4) into the expression above to get the value of the definite integral.