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

∫2cot^2 2t

$$\int\mathrm{2}{cot}^{\mathrm{2}} \mathrm{2}{t} \\ $$

Commented by Tikufly last updated on 01/Oct/17

I think your question should  be ∫2cot^2 2tdt

$$\mathrm{I}\:\mathrm{think}\:\mathrm{your}\:\mathrm{question}\:\mathrm{should} \\ $$$$\mathrm{be}\:\int\mathrm{2cot}^{\mathrm{2}} \mathrm{2tdt} \\ $$

Answered by alex041103 last updated on 01/Oct/17

∫2cot^2 (2t)dt  We make the substitution  u=2t, du=2dt  ∫2cot^2 (2t)dt=∫cot^2 (u)du=  =∫ ((cos^2 u)/(sin^2 u))du=∫cosu ((cosu)/(sin^2 u)) du=I  Now we integrate by parts  x=cosu    dv=((cosu)/(sin^2 u))du  dx=−sinu du   v=−(1/(sinu))  ⇒I=−((cosu)/(sinu))−∫ du=−cot(u)−u=  =−cot(2t)−2t+C  Ans. ∫2cot^2 (2t) dt=−cot(2t)−2t+C

$$\int\mathrm{2}{cot}^{\mathrm{2}} \left(\mathrm{2}{t}\right){dt} \\ $$$${We}\:{make}\:{the}\:{substitution} \\ $$$${u}=\mathrm{2}{t},\:{du}=\mathrm{2}{dt} \\ $$$$\int\mathrm{2}{cot}^{\mathrm{2}} \left(\mathrm{2}{t}\right){dt}=\int{cot}^{\mathrm{2}} \left({u}\right){du}= \\ $$$$=\int\:\frac{{cos}^{\mathrm{2}} {u}}{{sin}^{\mathrm{2}} {u}}{du}=\int{cosu}\:\frac{{cosu}}{{sin}^{\mathrm{2}} {u}}\:{du}={I} \\ $$$${Now}\:{we}\:{integrate}\:{by}\:{parts} \\ $$$${x}={cosu}\:\:\:\:{dv}=\frac{{cosu}}{{sin}^{\mathrm{2}} {u}}{du} \\ $$$${dx}=−{sinu}\:{du}\:\:\:{v}=−\frac{\mathrm{1}}{{sinu}} \\ $$$$\Rightarrow{I}=−\frac{{cosu}}{{sinu}}−\int\:{du}=−{cot}\left({u}\right)−{u}= \\ $$$$=−{cot}\left(\mathrm{2}{t}\right)−\mathrm{2}{t}+{C} \\ $$$${Ans}.\:\int\mathrm{2}{cot}^{\mathrm{2}} \left(\mathrm{2}{t}\right)\:{dt}=−{cot}\left(\mathrm{2}{t}\right)−\mathrm{2}{t}+{C} \\ $$

Answered by Tikufly last updated on 01/Oct/17

If your question is  ∫2cot^2 2tdt, then  I=∫2(cosec^2 2t−1)dt    =∫2cosec^2 2t−∫2dt    =−cot2t−2t+C

$$\mathrm{If}\:\mathrm{your}\:\mathrm{question}\:\mathrm{is} \\ $$$$\int\mathrm{2}{cot}^{\mathrm{2}} \mathrm{2}{tdt},\:{then} \\ $$$${I}=\int\mathrm{2}\left({cosec}^{\mathrm{2}} \mathrm{2}{t}−\mathrm{1}\right){dt} \\ $$$$\:\:=\int\mathrm{2}{cosec}^{\mathrm{2}} \mathrm{2}{t}−\int\mathrm{2}{dt} \\ $$$$\:\:=−{cot}\mathrm{2}{t}−\mathrm{2}{t}+{C} \\ $$

Commented by Isse last updated on 01/Oct/17

thnks

$${thnks} \\ $$

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