There is a direct u-sub to handle this integral, but the way you have done it does illustrate an important concept.LocalStudent said:When I did integration by parts I got to this:
tan^2 x - ∫(sec^2 x tanx) dx (integral also from 0 to ∏/3)
which take you back to the question I started with.
Jorriss said:Suppose, I = ∫Adx and you perform integration by parts and get I = B - ∫Adx. Well, that second integral is, as you noted, what you started with. Thus, I = B - I and so 2I = B and thus, I=B/2.
integrals.
Jorriss said:There is a direct u-sub to handle this integral, but the way you have done it does illustrate an important concept.
Suppose, I = ∫Adx and you perform integration by parts and get I = B - ∫Adx. Well, that second integral is, as you noted, what you started with. Thus, I = B - I and so 2I = B and thus, I=B/2.
That type of oscillatory behavior is very important for many trigonometric and exponential integrals.
The equation is ∫(sec^2 x tanx) dx from 0 to ∏/3.
To solve this integral, use the trigonometric identity sec^2 x = 1 + tan^2 x and then use substitution to change the integral into the form ∫(1 + tan^2 x) tanx dx. Then, use integration by parts to solve the integral.
The value of the integral is approximately 0.75.
First, use the trigonometric identity sec^2 x = 1 + tan^2 x to change the integral into the form ∫(1 + tan^2 x) tanx dx. Then, use u-substitution with u = tanx and du = sec^2 x dx to change the integral into the form ∫(1 + u^2) du. Use integration by parts with u = u and dv = 1 + u^2 to solve the integral. Finally, substitute back in u = tanx and solve for the limits of integration to get the final answer.
The limits of integration, 0 and ∏/3, determine the range of values for x in the integral. In this case, the integral is being solved from 0 to ∏/3, which means that the area under the curve is being calculated within that specific range. It is important to have accurate limits of integration to get the correct value for the integral.