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

Oct 5, 2017

${\int}_{0}^{\frac{\pi}{4}} \left[1 - \tan \left(4 x\right)\right] \mathrm{dx} = \frac{\pi}{4}$

#### Explanation:

Let $4 x = u$, then $4 \mathrm{dx} = \mathrm{du}$ and we can write ${\int}_{0}^{\frac{\pi}{4}} \left[1 - \tan \left(4 x\right)\right] \mathrm{dx}$ as

${\int}_{0}^{\pi} \left[1 - \tan u\right] \frac{\mathrm{du}}{4}$ - as limits are now $0 \times 4 = 0$ and $\frac{\pi}{4} \times 4 = \pi$

= $\frac{1}{4} {\int}_{0}^{\pi} \left[1 - \tan u\right] \mathrm{du}$

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

= $\frac{1}{4} \left[\pi + 0 - \left(0 + 0\right)\right]$ as both $| \cos \pi | = | \cos 0 | = 1$

= $\frac{\pi}{4}$