Use the derivatives of sinx and cosx to show that d/dx [tanx] = sec^2 and that d/dx [secx]=sec^2xsinx.
Hence evaluate ∫ [1 + sinx]/[cos^2x] dx limit from 0 to pi
I know the first part of the question, i'm not sure how i should do the second part "hence ..."
To show that the derivative of tan(x) is equal to sec^2(x), we start with the definitions of sin(x), cos(x), and tan(x):
sin(x) = opposite/hypotenuse
cos(x) = adjacent/hypotenuse
tan(x) = opposite/adjacent
Using these definitions, we can rewrite tan(x) as the ratio sin(x)/cos(x):
tan(x) = sin(x)/cos(x)
Now, to find the derivative of tan(x) with respect to x, we use the quotient rule:
d/dx(tan(x)) = (d/dx(sin(x))/cos(x) - sin(x)*d/dx(cos(x)))/cos^2(x)
We can differentiate sin(x) and cos(x) using their respective derivatives:
d/dx(sin(x)) = cos(x)
d/dx(cos(x)) = -sin(x)
Substituting these derivatives in, we have:
d/dx(tan(x)) = (cos(x)/cos(x) - sin(x)*(-sin(x)))/cos^2(x)
= (cos(x) + sin^2(x))/cos^2(x)
= (1 + sin^2(x))/cos^2(x)
But we know that sin^2(x) + cos^2(x) = 1, so we can rewrite it as:
d/dx(tan(x)) = 1/cos^2(x)
= sec^2(x)
Therefore, we have shown that the derivative of tan(x) is equal to sec^2(x).
Now let's move on to the second part of the question.
To evaluate the integral ∫ [1 + sin(x)]/[cos^2(x)] dx from 0 to pi, we can use the result we just derived: d/dx(sec(x)) = sec^2(x).
By rewriting the integrand, we have:
∫ [1 + sin(x)]/[cos^2(x)] dx
= ∫ 1/[cos^2(x)] dx + ∫ sin(x)/[cos^2(x)] dx
The first integral can be simplified using the derivative of sec(x). We have:
∫ 1/[cos^2(x)] dx = ∫ sec^2(x) dx
And we know from the derivative of sec(x) that d/dx(sec(x)) = sec^2(x). Therefore:
∫ sec^2(x) dx = sec(x) + C1
So the first integral becomes sec(x) + C1, where C1 is the constant of integration.
For the second integral, we can perform a substitution using u = cos(x), du = -sin(x) dx. This changes the integral to:
-∫ du/u^2
Integrating this, we get:
-∫ du/u^2 = 1/u + C2 = 1/cos(x) + C2 = sec(x) + C2
where C2 is another constant of integration.
Now, to evaluate the original integral from 0 to pi, we substitute the limits in:
[∫ [1 + sin(x)]/[cos^2(x)] dx] from 0 to pi
= [sec(x) + C1 - sec(x) - C2] from 0 to pi
= (sec(pi) + C1 - sec(pi) - C2) - (sec(0) + C1 - sec(0) - C2)
= (1 + C1 - 1 - C2) - (1 + C1 - 1 - C2)
= C1 - C2
Since C1 and C2 are arbitrary constants, their difference C1 - C2 is also a constant. Therefore, the result of the integral from 0 to pi is a constant value.