∫ tan^3 2xdx


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Question Number 116024 by mohammad17 last updated on 30/Sep/20

$\int {t a n}^{3} 2 x d x$
	Answered by bemath last updated on 30/Sep/20

	$= \frac{1}{2} \int \tan (2 x) (\sec^{2} 2 x - 1) d (2 x)$ $= \frac{1}{2} [\int \tan (2 x) d (\tan (2 x)] - \int \frac{\sin (2 x)}{\cos (2 x)} d (2 x)$ $= \frac{1}{2} [\frac{1}{2} \tan^{2} (2 x) + \ln (\cos (2 x))] + c$ $= \frac{1}{4} \tan^{2} (2 x) + \frac{1}{2} \ln (\cos (2 x)) + c$
	Answered by Dwaipayan Shikari last updated on 30/Sep/20

	$\int \tan^{2} 2 x \tan 2 xdx$ $\int (\sec^{2} 2 x - 1) \tan 2 xdx$ $\int \sec^{2} 2 xtan 2 xdx - \int \tan 2 xdx$ $\frac{1}{4} . \tan^{2} 2 x - \frac{1}{2} \log (\sec 2 x) dx$ $\frac{1}{4} \tan^{2} 2 x + \frac{1}{2} \log (\cos 2 x) + C$
	Answered by MJS_new last updated on 30/Sep/20

	$\int \tan^{3} 2 x d x =$ $[t = \tan^{3} 2 x \to d x = \frac{d t}{6 t^{3 / 2} (t^{3 / 2} + 1)}]$ $= \frac{1}{6} \int \frac{t^{1 / 3}}{t^{2 / 3} + 1} d t = \frac{1}{6} \int \frac{d t}{t^{1 / 3}} - \frac{1}{6} \int \frac{d t}{t^{1 / 3} (t^{2 / 3} + 1)} =$ $[u = t^{2 / 3} \to d t = \frac{3}{2} t^{1 / 3} d t in the 2^{nd} one]$ $= \frac{1}{4} t^{2 / 3} - \frac{1}{4} \ln (t^{2 / 3} + 1) =$ $= \frac{1}{4} \tan^{2} 2 x - \frac{1}{4} \ln (1 + \tan^{2} 2 x) + C$
	Answered by mnjuly1970 last updated on 30/Sep/20

	$2 x = t \Rightarrow d x = \frac{1}{2} d t$ $I = \frac{1}{2} \int {t a n}^{3} (t) d t = \frac{1}{2} \int [{t a n}^{3} (t) + t a n (t) - t a n n (t)] d t$ $= \frac{1}{2} \int t a n (t) (1 + {t a n}^{2} (t)) d t + \frac{1}{2} l n (c o s (t)) + C$ $= \frac{1}{4} {t a n}^{2} (t) + \frac{1}{2} l n (c o s (t)) + C$ $= \frac{1}{4} {t a n}^{2} (2 x) + \frac{1}{2} l n (c o s (2 x)) + C$
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