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Question Number 37432 by MJS last updated on 13/Jun/18
∫(dα/((1/2)tan α +2cot (α/2)))=? ∫(dβ/(2tan (β/2) +(1/2)cot β))=?
$$\int\frac{{d}\alpha}{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}\:\alpha\:+\mathrm{2cot}\:\frac{\alpha}{\mathrm{2}}}=? \\ $$$$\int\frac{{d}\beta}{\mathrm{2tan}\:\frac{\beta}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cot}\:\beta}=? \\ $$
Answered by ajfour last updated on 13/Jun/18
∫((2sin αcos αdα)/(sin^2 α+4cos^2 α))=∫((sin 2αdα)/(1+(3/2)(1+cos 2α))) =(1/3)∫ ((3sin 2α d(2α))/(5+3cos 2α)) =−(1/3)ln (5+3cos 2α)+c .
$$\int\frac{\mathrm{2sin}\:\alpha\mathrm{cos}\:\alpha{d}\alpha}{\mathrm{sin}\:^{\mathrm{2}} \alpha+\mathrm{4cos}\:^{\mathrm{2}} \alpha}=\int\frac{\mathrm{sin}\:\mathrm{2}\alpha{d}\alpha}{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\alpha\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\:\frac{\mathrm{3sin}\:\mathrm{2}\alpha\:{d}\left(\mathrm{2}\alpha\right)}{\mathrm{5}+\mathrm{3cos}\:\mathrm{2}\alpha} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\left(\mathrm{5}+\mathrm{3cos}\:\mathrm{2}\alpha\right)+{c}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jun/18
2)t=tan(β/2) dt=sec^2 (β/2)×(1/2)×dβ ∫((2dt)/(1+t^2 )) ×(1/(2t+(1/2)×((1−t^2 )/(2t)))) ∫((2dt)/(1+t^2 ))×((4t)/(8t^2 +1−t^2 )) 8∫((tdt)/(1+t^2 ))×(1/(1+7t^2 )) k=t^2 dk=2tdt 4∫(dk/(1+k))×(1/(1+7k)) (4/6)∫(((7+7k)−(1+7k))/((1+k)(1+7k)))dk (2/3)∫((7dk)/(1+7k))−(2/3)∫(dk/(1+k)) (2/3)ln∣((1+7k)/(1+k))∣ (2/3)ln∣((1+7t^2 )/(1+t^2 ))∣ (2/3)ln∣((1+7tan^2 β)/(1+tan^2 β))∣
$$\left.\mathrm{2}\right){t}={tan}\frac{\beta}{\mathrm{2}}\:\:\:{dt}={sec}^{\mathrm{2}} \frac{\beta}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}×{d}\beta \\ $$$$\int\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:×\frac{\mathrm{1}}{\mathrm{2}{t}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}{t}}} \\ $$$$\int\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }×\frac{\mathrm{4}{t}}{\mathrm{8}{t}^{\mathrm{2}} +\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$\mathrm{8}\int\frac{{tdt}}{\mathrm{1}+{t}^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{1}+\mathrm{7}{t}^{\mathrm{2}} }\:\:\:\:{k}={t}^{\mathrm{2}} \:\:\:\:\:{dk}=\mathrm{2}{tdt} \\ $$$$\mathrm{4}\int\frac{{dk}}{\mathrm{1}+{k}}×\frac{\mathrm{1}}{\mathrm{1}+\mathrm{7}{k}} \\ $$$$\frac{\mathrm{4}}{\mathrm{6}}\int\frac{\left(\mathrm{7}+\mathrm{7}{k}\right)−\left(\mathrm{1}+\mathrm{7}{k}\right)}{\left(\mathrm{1}+{k}\right)\left(\mathrm{1}+\mathrm{7}{k}\right)}{dk} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\int\frac{\mathrm{7}{dk}}{\mathrm{1}+\mathrm{7}{k}}−\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{dk}}{\mathrm{1}+{k}} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}{ln}\mid\frac{\mathrm{1}+\mathrm{7}{k}}{\mathrm{1}+{k}}\mid \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}{ln}\mid\frac{\mathrm{1}+\mathrm{7}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\mid \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}{ln}\mid\frac{\mathrm{1}+\mathrm{7}{tan}^{\mathrm{2}} \beta}{\mathrm{1}+{tan}^{\mathrm{2}} \beta}\mid \\ $$$$ \\ $$$$ \\ $$

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