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Question-203693




Question Number 203693 by Numsey last updated on 26/Jan/24
Answered by MM42 last updated on 26/Jan/24
I=∫(√2)cos(x/2)dx=2(√2)sin(x/2) +c
$${I}=\int\sqrt{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}{dx}=\mathrm{2}\sqrt{\mathrm{2}}{sin}\frac{{x}}{\mathrm{2}}\:+{c} \\ $$
Commented by Frix last updated on 26/Jan/24
(√(1+cos x))=(√2)∣cos (x/2)∣
$$\sqrt{\mathrm{1}+\mathrm{cos}\:{x}}=\sqrt{\mathrm{2}}\mid\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\mid \\ $$
Answered by esmaeil last updated on 26/Jan/24
I=2(√2)∣sin(x/2)∣+c  DomainI=[0,+∞)
$${I}=\mathrm{2}\sqrt{\mathrm{2}}\mid{sin}\frac{{x}}{\mathrm{2}}\mid+{c} \\ $$$${DomainI}=\left[\mathrm{0},+\infty\right) \\ $$$$ \\ $$

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