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Question Number 190937 by Spillover last updated on 14/Apr/23
Show  that                       ∫  ((sech (√x) tanh (√x))/( (√x)))=−(2/(cosh (√x)))
$$\mathrm{Show}\:\:\mathrm{that}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\:\frac{\mathrm{sech}\:\sqrt{\mathrm{x}}\:\mathrm{tanh}\:\sqrt{\mathrm{x}}}{\:\sqrt{\mathrm{x}}}=−\frac{\mathrm{2}}{\mathrm{cosh}\:\sqrt{\mathrm{x}}} \\ $$
Answered by ARUNG_Brandon_MBU last updated on 15/Apr/23
I=∫(((sech(√x))(tanh(√x)))/( (√x)))dx=∫((sinh(√x))/( (√x)cosh^2 (√x)))dx  t=cosh(√x) ⇒dt=((sinh(√x))/(2(√x)))dx  I=∫(2/t^2 )dt=−(2/t)+C=−(2/(cosh(√x)))+C
$${I}=\int\frac{\left(\mathrm{sech}\sqrt{{x}}\right)\left(\mathrm{tanh}\sqrt{{x}}\right)}{\:\sqrt{{x}}}{dx}=\int\frac{\mathrm{sinh}\sqrt{{x}}}{\:\sqrt{{x}}\mathrm{cosh}^{\mathrm{2}} \sqrt{{x}}}{dx} \\ $$$${t}=\mathrm{cosh}\sqrt{{x}}\:\Rightarrow{dt}=\frac{\mathrm{sinh}\sqrt{{x}}}{\mathrm{2}\sqrt{{x}}}{dx} \\ $$$${I}=\int\frac{\mathrm{2}}{{t}^{\mathrm{2}} }{dt}=−\frac{\mathrm{2}}{{t}}+{C}=−\frac{\mathrm{2}}{\mathrm{cosh}\sqrt{{x}}}+{C} \\ $$
Commented by Spillover last updated on 15/Apr/23
great
$$\mathrm{great} \\ $$

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