evaluate the integral:
integral from -pi/4 to 0 for the function 6sec^3x dx.
it has to be an exact answer and i did it and keep getting it wrong. I got
4sqrt(2)-4ln(-sqrt(2)+1)
According to the Wolfram integrator , this looks like a messy integration
http://integrals.wolfram.com/index.jsp?expr=6%2F%28cos%28x%29%29%5E3&random=false
looks like multiple integration by parts, arhhh!
u = sec x,
dv = sec^2 x dx
du = secx tanx dx
v = tan x
∫sec^3 x dx
= ∫u dv = uv - ∫v du
= secx tanx - ∫secx tan^2 x dx
= secx tanx - ∫(secx (sec^2 x - 1) dx
= secx tanx - ∫sec^3 x dx + ∫secx dx
so,
2∫sec^3 x dx = secx tanx + ∫secx dx
= secx tanx + ln(secx tanx)
∫sec^3 x dx = 1/2 (secx tanx + ln(secx + tanx)
plug in 0 and pi/4 to get
1/2 (√2*1 + ln(√2+1)) - 1/2(1*0 + ln(1+0))
= 1/2 (√2 + ln(√2+1))
multiply by 6 to get 3(√2 + ln(√2+1))
double check my math, and you will either
(a) see your mistake
(b) see my mistake
To evaluate the integral of the function 6sec^3x dx from -pi/4 to 0, we can use the following steps:
Step 1: Rewrite the integral using the identity sec^2(x) = 1 + tan^2(x):
∫6sec^3(x) dx = ∫6sec(x)sec^2(x) dx
= ∫6sec(x)(1 + tan^2(x)) dx
Step 2: Apply the substitution method:
Let u = tan(x), then du = sec^2(x) dx
Step 3: Substitute back into the integral:
∫6sec(x)(1 + tan^2(x)) dx = ∫6sec(x)(1 + u^2) dx
Step 4: Evaluate the new integral:
Now, the integral becomes: ∫6sec(x)(1 + u^2) dx = 6 ∫(sec(x) + u^2sec(x)) dx
We can evaluate each term separately:
∫sec(x) dx = ln|sec(x) + tan(x)| + C
∫u^2sec(x) dx = u^2ln|sec(x) + tan(x)| - 2∫u(u'ln|sec(x) + tan(x)|) dx
= u^2ln|sec(x) + tan(x)| - 2∫u(u') dx
= u^2ln|sec(x) + tan(x)| - 2∫u du
= u^2ln|sec(x) + tan(x)| - u^2 + C
Therefore, the full integral is:
6( ln|sec(x) + tan(x)| + u^2ln|sec(x) + tan(x)| - u^2 ) + C
= 6[ ln|sec(x) + tan(x)| + tan^2(x)ln|sec(x) + tan(x)| - tan^2(x) ] + C
Step 5: Substitute the original variable back in:
Since u = tan(x), we have:
= 6[ ln|sec(x) + tan(x)| + tan^2(x)ln|sec(x) + tan(x)| - tan^2(x) ] + C
= 6[ ln|sec(x) + tan(x)| + tan^2(x)ln|sec(x) + tan(x)| - tan^2(x) ] + C
Step 6: Evaluate the integral over the given limits:
Now, we can substitute the upper limit (0) and the lower limit (-pi/4) into the expression and subtract the results to find the exact answer.
= 6[ ln|sec(0) + tan(0)| + tan^2(0)ln|sec(0) + tan(0)| - tan^2(0) ]
- 6[ ln|sec(-pi/4) + tan(-pi/4)| + tan^2(-pi/4)ln|sec(-pi/4) + tan(-pi/4)| - tan^2(-pi/4) ]
= 6[ ln|1 + 0| + (0)ln|1 + 0| - (0) ]
- 6[ ln|1 + (-1)| + (-1^2)ln|1 + (-1)| - (-1^2) ]
= 6[ ln(1) + 0 - 0 ]
- 6[ ln(0) + 1*ln(0) + 1 ]
= 6( 0 ) - 6( undefined ) = 0 - undefined = undefined
Therefore, the integral from -pi/4 to 0 for the function 6sec^3x dx is undefined.