Evaluate the integral.[0,pi/4]∫( sec(t)tan(t)+ tcos(2t)j+ ((sin(2t))^2) *(cos(2t))k) dt
do the j and k integrals separately
∫sect tant dt = sect
∫t cos(2t) dt: use integration by parts, with u=t, dv = cos(2t) dt
∫sin^2(2t) dt = ∫(1-cos(4t))/2 dt
see what you can do with that. You can always check your work at wolframalpha.com.
To evaluate the given integral, we will break it down into three separate integrals for each of the components: the first component involving sec(t)tan(t), the second component involving tcos(2t)j, and the third component involving ((sin(2t))^2) *(cos(2t))k.
Let's start by evaluating the integral of sec(t)tan(t):
∫(sec(t)tan(t)) dt
To solve this, we can use a substitution. Let's substitute u = sec(t), which means du = sec(t)tan(t) dt.
Now, our integral becomes:
∫ du
The integral of du is simply the variable u:
u = sec(t)
Next, let's evaluate the integral of tcos(2t)j:
∫(tcos(2t)) dt
We can solve this integral by using integration by parts. Integration by parts states that if we have the product of two functions, u and v, then ∫(u dv) = u v - ∫(v du).
In this case, let's choose:
u = t (differentiate to get du)
v = sin(2t) (integrate to get dv)
Applying integration by parts formula, we have:
∫(tcos(2t)) dt = t(-1/2)sin(2t) - ∫((-1/2)sin(2t) dt)
Simplifying further, we get:
∫(tcos(2t)) dt = -1/2 t sin(2t) + ∫(1/2sin(2t)) dt
Using integration by parts once again, we have:
∫(tcos(2t)) dt = -1/2 t sin(2t) + (-1/4) cos(2t)
Finally, let's evaluate the integral of ((sin(2t))^2) *(cos(2t))k:
∫ ( ((sin(2t))^2) *(cos(2t)) ) dt
We will use the method of substitution again for this integral. Let's substitute u = sin(2t), which means du = 2cos(2t) dt.
Now, our integral becomes:
∫ ( u^2 ) du / 2
Simplifying further, we get:
(1/2) ∫ ( u^2 ) du
The integral of u^2 is (u^3)/3:
(1/2) (u^3)/3 = (u^3)/6
Substituting back u = sin(2t), we have:
( (sin(2t))^3 )/6
Now that we have evaluated all three integrals, let's put them together:
[0,pi/4]∫ ( sec(t)tan(t) + tcos(2t)j + ((sin(2t))^2) *(cos(2t))k ) dt
= [0,pi/4] ( sec(t)tan(t) ) dt + [0,pi/4] ( -1/2 t sin(2t) + (-1/4) cos(2t) ) dt + [0,pi/4] ( (sin(2t))^3 )/6 dt
We can now evaluate each of these integrals individually using the antiderivative rules and the given limits [0, pi/4].