integration of tan4x
answer for integration of tan4x..
is it \(\large \tan^4x\) or \(\large \tan4x\) ?
hullo? +_+
Its tan raise to 4x.... ;
\[\large \int\limits \tan^4x\;dx \qquad = \qquad \int\limits \tan^2x \cdot \tan^2x \;dx\]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,\[\large \int\limits\limits \tan^2x \cdot \color{orangered}{(\sec^2x-1)} \;dx \qquad = \qquad \int\limits \tan^2x \sec^2x \;dx-\int\limits \tan^2x\;dx\]
Lemme know if you're confused by any of that so far.
Thank you so much sir... : )
\[\large \int\limits\limits \tan^2x \sec^2x \;dx-\int\limits\limits \tan^2x\;dx\]For the first integral, apply a `u substitution`, let \(\large u=\tan x\). For the second integral, apply the trig identity again,\[\large \int\limits\limits\limits u^2 \sec^2x \;dx-\int\limits\limits\limits \sec^2x-1\;dx\]
Is this the final answer.
no c:
just showing you some of the steps. Do you understand how to apply a u sub to the first one? :o
ya.......
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