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

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Answer 1 · 2017-10-05T16:05:01

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

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 $|cospi|=|cos0|=1$

= $pi/4$

Find the integral int_0^(pi/4) [1-tan(4x)]dxint_0^(pi/4) [1-tan(4x)]dx?