∫2cot^2 2t


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Question Number 21701 by Isse last updated on 01/Oct/17

$\int 2 {c o t}^{2} 2 t$
Commented by Tikufly last updated on 01/Oct/17

$I think your question should$ $be \int {2 \cot}^{2} 2 tdt$
	Answered by alex041103 last updated on 01/Oct/17

	$\int 2 {c o t}^{2} (2 t) d t$ $W e m a k e t h e s u b s t i t u t i o n$ $u = 2 t, d u = 2 d t$ $\int 2 {c o t}^{2} (2 t) d t = \int {c o t}^{2} (u) d u =$ $= \int \frac{{c o s}^{2} u}{{s i n}^{2} u} d u = \int c o s u \frac{c o s u}{{s i n}^{2} u} d u = I$ $N o w w e i n t e g r a t e b y p a r t s$ $x = c o s u d v = \frac{c o s u}{{s i n}^{2} u} d u$ $d x = - s i n u d u v = - \frac{1}{s i n u}$ $\Rightarrow I = - \frac{c o s u}{s i n u} - \int d u = - c o t (u) - u =$ $= - c o t (2 t) - 2 t + C$ $A n s . \int 2 {c o t}^{2} (2 t) d t = - c o t (2 t) - 2 t + C$
	Answered by Tikufly last updated on 01/Oct/17

	$If your question is$ $\int 2 {c o t}^{2} 2 t d t, t h e n$ $I = \int 2 ({c o s e c}^{2} 2 t - 1) d t$ $= \int 2 {c o s e c}^{2} 2 t - \int 2 d t$ $= - c o t 2 t - 2 t + C$
	Commented by Isse last updated on 01/Oct/17

	$t h n k s$
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