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Question Number 206536 by cortano21 last updated on 18/Apr/24

⋖ ♠[2^x sin ((√(4−2^(x+2) )) )

$$\:\:\:\:\underline{\underbrace{\lessdot}\cancel{} }\underbrace{\nsupseteqq\spadesuit\left[}\mathrm{2}^{{x}} \:\mathrm{sin}\:\left(\sqrt{\mathrm{4}−\mathrm{2}^{{x}+\mathrm{2}} }\:\right)\right. \\ $$

Commented by Frix last updated on 18/Apr/24

Simply use t=(√(1−2^x ))

$$\mathrm{Simply}\:\mathrm{use}\:{t}=\sqrt{\mathrm{1}−\mathrm{2}^{{x}} } \\ $$

Answered by lepuissantcedricjunior last updated on 20/Apr/24

I=∫((2^x sin((√(4−2^(x+2) ))))/( (√(1−2^x ))cos((√(1−2^x )))))dx =∫((2^x sin(2(√(1−2^x ))))/( (√(1−2^x ))cos((√(1−2^x )))))dx =∫((2^(x+1) sin((√(1−2^x ))))/( (√(1−2^x ))))dx posons 1−2^x =t^2 =>2^x =1−t^2 =>ln2×2^x dx=−2tdt ⇒dx=−((2t)/((1−t^2 )ln2)) dt ⇔I=∫((−4t(1−t^2 )sint)/(tln2(1−t^2 )))dt =∫−4((sint)/(ln(2)))dt=4((cost)/(ln2))+c c∈R ⇔I=((4cos((√(1−2^x ))))/(ln2))+c c∈R

$$\boldsymbol{{I}}=\int\frac{\mathrm{2}^{\boldsymbol{{x}}} \boldsymbol{{sin}}\left(\sqrt{\mathrm{4}−\mathrm{2}^{\boldsymbol{{x}}+\mathrm{2}} }\right)}{\:\sqrt{\mathrm{1}−\mathrm{2}^{\boldsymbol{{x}}} }\boldsymbol{{cos}}\left(\sqrt{\mathrm{1}−\mathrm{2}^{\boldsymbol{{x}}} }\right)}\boldsymbol{{dx}} \\ $$$$\:\:=\int\frac{\mathrm{2}^{\boldsymbol{{x}}} \boldsymbol{{sin}}\left(\mathrm{2}\sqrt{\mathrm{1}−\mathrm{2}^{\boldsymbol{{x}}} }\right)}{\:\sqrt{\mathrm{1}−\mathrm{2}^{\boldsymbol{{x}}} }\boldsymbol{{cos}}\left(\sqrt{\mathrm{1}−\mathrm{2}^{\boldsymbol{{x}}} }\right)}\boldsymbol{{dx}} \\ $$$$\:\:=\int\frac{\mathrm{2}^{\boldsymbol{{x}}+\mathrm{1}} \boldsymbol{{sin}}\left(\sqrt{\mathrm{1}−\mathrm{2}^{\boldsymbol{{x}}} }\right)}{\:\sqrt{\mathrm{1}−\mathrm{2}^{\boldsymbol{{x}}} }}\boldsymbol{{dx}} \\ $$$$\:\:\boldsymbol{{posons}}\:\mathrm{1}−\mathrm{2}^{\boldsymbol{{x}}} =\boldsymbol{{t}}^{\mathrm{2}} =>\mathrm{2}^{\boldsymbol{{x}}} =\mathrm{1}−\boldsymbol{{t}}^{\mathrm{2}} \\ $$$$=>\boldsymbol{{ln}}\mathrm{2}×\mathrm{2}^{\boldsymbol{{x}}} \boldsymbol{{dx}}=−\mathrm{2}\boldsymbol{{tdt}} \\ $$$$\Rightarrow\boldsymbol{{dx}}=−\frac{\mathrm{2}\boldsymbol{{t}}}{\left(\mathrm{1}−\boldsymbol{{t}}^{\mathrm{2}} \right)\boldsymbol{{ln}}\mathrm{2}}\:\boldsymbol{{dt}} \\ $$$$\Leftrightarrow\boldsymbol{{I}}=\int\frac{−\mathrm{4}\boldsymbol{{t}}\left(\mathrm{1}−\boldsymbol{{t}}^{\mathrm{2}} \right)\boldsymbol{{sint}}}{\boldsymbol{{tln}}\mathrm{2}\left(\mathrm{1}−\boldsymbol{{t}}^{\mathrm{2}} \right)}\boldsymbol{{dt}} \\ $$$$\:\:\:\:\:\:\:=\int−\mathrm{4}\frac{\boldsymbol{{sint}}}{\boldsymbol{{ln}}\left(\mathrm{2}\right)}\boldsymbol{{dt}}=\mathrm{4}\frac{\boldsymbol{{cost}}}{\boldsymbol{{ln}}\mathrm{2}}+\boldsymbol{{c}}\:\:\boldsymbol{{c}}\in\mathbb{R} \\ $$$$\Leftrightarrow\boldsymbol{{I}}=\frac{\mathrm{4}\boldsymbol{{cos}}\left(\sqrt{\mathrm{1}−\mathrm{2}^{\boldsymbol{{x}}} }\right)}{\boldsymbol{{ln}}\mathrm{2}}+\boldsymbol{{c}}\:\:\:\:\boldsymbol{{c}}\in\mathbb{R} \\ $$$$ \\ $$

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