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

1 Answer
Shwetank Mauria
Oct 5, 2017

#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#

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