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Mathematics 16 Online
OpenStudy (anonymous):

integration of tan4x

OpenStudy (anonymous):

answer for integration of tan4x..

zepdrix (zepdrix):

is it \(\large \tan^4x\) or \(\large \tan4x\) ?

zepdrix (zepdrix):

hullo? +_+

OpenStudy (anonymous):

Its tan raise to 4x.... ;

zepdrix (zepdrix):

\[\large \int\limits \tan^4x\;dx \qquad = \qquad \int\limits \tan^2x \cdot \tan^2x \;dx\]Using our trig identity,\[\large \color{orangered}{\tan^2x+1=\sec^2x} \qquad \rightarrow \qquad \color{orangered}{\tan^2x=\sec^2x-1}\]Will give us,\[\large \int\limits\limits \tan^2x \cdot \color{orangered}{(\sec^2x-1)} \;dx \qquad = \qquad \int\limits \tan^2x \sec^2x \;dx-\int\limits \tan^2x\;dx\]

zepdrix (zepdrix):

Lemme know if you're confused by any of that so far.

OpenStudy (anonymous):

Thank you so much sir... : )

zepdrix (zepdrix):

\[\large \int\limits\limits \tan^2x \sec^2x \;dx-\int\limits\limits \tan^2x\;dx\]For the first integral, apply a `u substitution`, let \(\large u=\tan x\). For the second integral, apply the trig identity again,\[\large \int\limits\limits\limits u^2 \sec^2x \;dx-\int\limits\limits\limits \sec^2x-1\;dx\]

OpenStudy (anonymous):

Is this the final answer.

zepdrix (zepdrix):

no c:

zepdrix (zepdrix):

just showing you some of the steps. Do you understand how to apply a u sub to the first one? :o

OpenStudy (anonymous):

ya.......

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