Question

What is the definite integral of `sec^4 x` from 0 to `pi/4`?

Answer 1

`int_0^(pi/4)sec^4(x)dx=int_0^(pi/4)sec^2(x)sec^2(x)dx`

Trig Identity

`sec^2(x)=tan^2(x)+1`

Use this identity to substitute for one of the `sec^2(x)`.

`int_0^(pi/4)[tan^2(x)+1]sec^2(x)dx`

Now begin with u-substitution

Let `u=tan(x)`

`du=sec^2(x) dx`

`int[u^2+1] du`

`[u^3/3+u]` Convert back to terms of x `-> [tan(x)^3/3+tan(x)]_0^(pi/4)`

`=[tan(pi/4)^3/3+tan(pi/4)-(tan(0)/3+tan(0))]`

`=[(1)^3/3+1-(0+0)]`

`=[1/3+1]`

`=[1/3+3/3]`

`=[4/3]`

`=1.3333`

Video solution here

Remember that, `sec(x)=1/cos(x)`

So we can also say, `sec^4(x)=1/(cos^4(x))`

After graphing press 2nd and then TRACE

Press 7 for integration, `intf(x)dx`

Then enter the LOWER and UPPER LIMITS

See the results of those actions below.