int_0^(pi/4)sec^4(x)dx=int_0^(pi/4)sec^2(x)sec^2(x)dx∫π40sec4(x)dx=∫π40sec2(x)sec2(x)dx
Trig Identity
sec^2(x)=tan^2(x)+1sec2(x)=tan2(x)+1
Use this identity to substitute for one of the sec^2(x)sec2(x).
int_0^(pi/4)[tan^2(x)+1]sec^2(x)dx∫π40[tan2(x)+1]sec2(x)dx
Now begin with u-substitution
Let u=tan(x)u=tan(x)
du=sec^2(x) dxdu=sec2(x)dx
int[u^2+1] du∫[u2+1]du
[u^3/3+u][u33+u] Convert back to terms of x -> [tan(x)^3/3+tan(x)]_0^(pi/4)→[tan(x)33+tan(x)]π40
=[tan(pi/4)^3/3+tan(pi/4)-(tan(0)/3+tan(0))]=⎡⎣tan(π4)33+tan(π4)−(tan(0)3+tan(0))⎤⎦
=[(1)^3/3+1-(0+0)]=[(1)33+1−(0+0)]
=[1/3+1]=[13+1]
=[1/3+3/3]=[13+33]
=[4/3]=[43]
=1.3333=1.3333
Video solution here
Remember that, sec(x)=1/cos(x)sec(x)=1cos(x)
So we can also say, sec^4(x)=1/(cos^4(x))sec4(x)=1cos4(x)
![enter image source here]()
After graphing press 2nd and then TRACE
Press 7 for integration, intf(x)dx∫f(x)dx
Then enter the LOWER and UPPER LIMITS
See the results of those actions below.
![enter image source here]()
