Evaluate the definite integral
(low) (high)
-π/6 ∫ π/6 2sec(x)tan(x) dx
hint:
What is the derivative of secx ??
this is one of the standard trig integrals. Recall that
d/dx sec(x) = sec(x) tan(x)
The integral should now be clear.
To evaluate the definite integral ∫(low)^(high) 2sec(x)tan(x) dx, where the limits of integration are -π/6 and π/6, we can follow these steps:
Step 1: Simplify the integrand
We can simplify the integrand 2sec(x)tan(x) using trigonometric identities.
The identity sec(x) = 1/cos(x) and tan(x) = sin(x)/cos(x), therefore we have:
2sec(x)tan(x) = 2(1/cos(x)) * (sin(x)/cos(x)) = 2sin(x)
So now our integrand simplifies to 2sin(x).
Step 2: Integrate the simplified function
The integral of sin(x) with respect to x is -cos(x).
Therefore, the definite integral becomes:
∫(-π/6)^(π/6) 2sec(x)tan(x) dx = ∫(-π/6)^(π/6) 2sin(x) dx = [-2cos(x)] evaluated from -π/6 to π/6.
Step 3: Evaluate the definite integral
Now we substitute the limits of integration into the antiderivative and subtract the value at the lower limit from the value at the upper limit:
[-2cos(x)] from -π/6 to π/6
= [-2cos(π/6)] - [-2cos(-π/6)]
= [-2√3/2] - [-2√3/2]
= -2√3/2 + 2√3/2
= 0
Therefore, the value of the definite integral ∫(-π/6)^(π/6) 2sec(x)tan(x) dx is 0.