Question

Find the integral `int_0^(pi/4) [1-tan(4x)]dx`?

Answer 1

`int_0^(pi/4) [1-tan(4x)]dx=pi/4`

Explanation:

Let `4x=u`, then `4dx=du` and we can write `int_0^(pi/4) [1-tan(4x)]dx` as

`int_0^pi [1-tanu] (du)/4` - as limits are now `0xx4=0` and `pi/4xx4=pi`

= `1/4int_0^pi [1-tanu]du`

= `1/4[u-(-ln|cosu|]_0^pi`

= `1/4[pi+0-(0+0)]` as both `|cospi|=|cos0|=1`

= `pi/4`