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Question Number 175531 by cortano1 last updated on 01/Sep/22
∫ (dt/(5cos t+6sin t)) =?
$$\:\int\:\frac{{dt}}{\mathrm{5cos}\:{t}+\mathrm{6sin}\:{t}}\:=? \\ $$
Answered by Ar Brandon last updated on 01/Sep/22
I=∫(dt/(5cost+6sint)) , x=tan(t/2) =∫(2/(5(((1−x^2 )/(1+x^2 )))+6(((2x)/(1+x^2 )))))∙(dx/(1+x^2 )) =∫((2dx)/(5+12x−5x^2 ))=∫((2dx)/(((61)/5)−5(x−(6/5))^2 )) =(2/( (√(61))))argtanh(((5x−6)/( (√(61)))))+C =(1/( (√(61))))ln∣((5x+(√(61))−6)/(5x−(√(61))−6))∣+C =(1/( (√(61))))ln∣((5tan((t/2))+(√(61))−6)/(5tan((t/2))−(√(61))−6))∣+C
$${I}=\int\frac{{dt}}{\mathrm{5cos}{t}+\mathrm{6sin}{t}}\:,\:{x}=\mathrm{tan}\frac{{t}}{\mathrm{2}} \\ $$$$\:\:\:=\int\frac{\mathrm{2}}{\mathrm{5}\left(\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\right)+\mathrm{6}\left(\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)}\centerdot\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\:\:\:=\int\frac{\mathrm{2}{dx}}{\mathrm{5}+\mathrm{12}{x}−\mathrm{5}{x}^{\mathrm{2}} }=\int\frac{\mathrm{2}{dx}}{\frac{\mathrm{61}}{\mathrm{5}}−\mathrm{5}\left({x}−\frac{\mathrm{6}}{\mathrm{5}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{61}}}\mathrm{argtanh}\left(\frac{\mathrm{5}{x}−\mathrm{6}}{\:\sqrt{\mathrm{61}}}\right)+{C} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{61}}}\mathrm{ln}\mid\frac{\mathrm{5}{x}+\sqrt{\mathrm{61}}−\mathrm{6}}{\mathrm{5}{x}−\sqrt{\mathrm{61}}−\mathrm{6}}\mid+{C} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{61}}}\mathrm{ln}\mid\frac{\mathrm{5tan}\left(\frac{{t}}{\mathrm{2}}\right)+\sqrt{\mathrm{61}}−\mathrm{6}}{\mathrm{5tan}\left(\frac{{t}}{\mathrm{2}}\right)−\sqrt{\mathrm{61}}−\mathrm{6}}\mid+{C} \\ $$
Commented by cortano1 last updated on 02/Sep/22
ok
$$\mathrm{ok} \\ $$
Answered by blackmamba last updated on 01/Sep/22
⇔ 5cos t+6sin t=(√(61)) ((5/( (√(61)))) cos t+(6/( (√(61))))sin t) = (√(61)) sin (α+t) I=∫ (dt/( (√(61)) sin (α+t))) I=(1/( (√(61)))) ln ∣ ((1−cos (α+t))/(sin (α+t)))∣ + c I=(1/( (√(61)))) ln ∣((1−((6/( (√(61)))) cos t−(5/( (√(61)))) sin t))/((5/( (√(61)))) cos t+(6/( (√(61)))) sin t)) ∣+c I=(1/( (√(61)))) ln ∣(((√(61))−6cos t+5sin t)/(5cos t+6sin t)) ∣ + c
$$\Leftrightarrow\:\mathrm{5cos}\:{t}+\mathrm{6sin}\:{t}=\sqrt{\mathrm{61}}\:\left(\frac{\mathrm{5}}{\:\sqrt{\mathrm{61}}}\:\mathrm{cos}\:{t}+\frac{\mathrm{6}}{\:\sqrt{\mathrm{61}}}\mathrm{sin}\:{t}\right) \\ $$$$\:\:\:\:\:\:=\:\sqrt{\mathrm{61}}\:\mathrm{sin}\:\left(\alpha+{t}\right) \\ $$$${I}=\int\:\frac{{dt}}{\:\sqrt{\mathrm{61}}\:\mathrm{sin}\:\left(\alpha+{t}\right)} \\ $$$${I}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{61}}}\:\mathrm{ln}\:\mid\:\frac{\mathrm{1}−\mathrm{cos}\:\left(\alpha+{t}\right)}{\mathrm{sin}\:\left(\alpha+{t}\right)}\mid\:+\:{c} \\ $$$${I}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{61}}}\:\mathrm{ln}\:\mid\frac{\mathrm{1}−\left(\frac{\mathrm{6}}{\:\sqrt{\mathrm{61}}}\:\mathrm{cos}\:{t}−\frac{\mathrm{5}}{\:\sqrt{\mathrm{61}}}\:\mathrm{sin}\:{t}\right)}{\frac{\mathrm{5}}{\:\sqrt{\mathrm{61}}}\:\mathrm{cos}\:{t}+\frac{\mathrm{6}}{\:\sqrt{\mathrm{61}}}\:\mathrm{sin}\:{t}}\:\mid+{c} \\ $$$${I}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{61}}}\:\mathrm{ln}\:\mid\frac{\sqrt{\mathrm{61}}−\mathrm{6cos}\:{t}+\mathrm{5sin}\:{t}}{\mathrm{5cos}\:{t}+\mathrm{6sin}\:{t}}\:\mid\:+\:{c}\: \\ $$
Commented by cortano1 last updated on 02/Sep/22
ok
$$\mathrm{ok} \\ $$

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