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Question Number 145385 by mnjuly1970 last updated on 04/Jul/21

∫_0 ^( (π/2)) ((1+cos (2x))/(sin (2x ))). ln((sec (x)))^(1/3) dx=?

$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+\mathrm{cos}\:\left(\mathrm{2x}\right)}{\mathrm{sin}\:\left(\mathrm{2x}\:\right)}.\:\mathrm{ln}\sqrt[{\mathrm{3}}]{\mathrm{sec}\:\left(\mathrm{x}\right)}\:\mathrm{dx}=? \\ $$

Answered by mindispower last updated on 04/Jul/21

cos(2x)=2cos^2 (x)−1 ⇔−(1/3)∫_0 ^(π/2) ((cos(x))/(sin(x)))ln(cos(x))dx =−(1/(12)).∂_a 2∫_0 ^(π/2) cos^(2a−1) (x)sin^(2b−1) (x)dx∣_((a,b)=((3/2),0)) =−lim_(b→0) (1/(12))∂_a β(a,b)∣(a,b)=((3/2),b) =−lim_(b→0) (1/(12))β(a,b)(Ψ(a)−Ψ(a+b)) =−(1/(12)) lim_(b→0) Γ(b)(Ψ((3/2))−Ψ((3/2)+b) =−(1/(12))(Ψ((3/2))−Ψ((3/2)+b))Γ(b) Ψ(a)−Ψ(a+h)=−Ψ′(a)h Ψ((3/2))−Ψ((3/2)+b)=−Ψ′((3/2))b ⇔(1/(12))lim_(b→0) bΓ(b)Ψ′((3/2))=(1/(12))Γ(1+b)Ψ′((3/2))=((Ψ′((3/2)))/(12))

$${cos}\left(\mathrm{2}{x}\right)=\mathrm{2}{cos}^{\mathrm{2}} \left({x}\right)−\mathrm{1} \\ $$$$\Leftrightarrow−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}\left({x}\right)}{{sin}\left({x}\right)}{ln}\left({cos}\left({x}\right)\right){dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{12}}.\partial_{{a}} \mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}{a}−\mathrm{1}} \left({x}\right){sin}^{\mathrm{2}{b}−\mathrm{1}} \left({x}\right){dx}\mid_{\left({a},{b}\right)=\left(\frac{\mathrm{3}}{\mathrm{2}},\mathrm{0}\right)} \\ $$$$=−\underset{{b}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{12}}\partial_{{a}} \beta\left({a},{b}\right)\mid\left({a},{b}\right)=\left(\frac{\mathrm{3}}{\mathrm{2}},{b}\right) \\ $$$$=−\underset{{b}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{12}}\beta\left({a},{b}\right)\left(\Psi\left({a}\right)−\Psi\left({a}+{b}\right)\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{12}}\:\underset{{b}\rightarrow\mathrm{0}} {\mathrm{lim}}\Gamma\left({b}\right)\left(\Psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\Psi\left(\frac{\mathrm{3}}{\mathrm{2}}+{b}\right)\right. \\ $$$$=−\frac{\mathrm{1}}{\mathrm{12}}\left(\Psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\Psi\left(\frac{\mathrm{3}}{\mathrm{2}}+{b}\right)\right)\Gamma\left({b}\right) \\ $$$$\Psi\left({a}\right)−\Psi\left({a}+{h}\right)=−\Psi'\left({a}\right){h} \\ $$$$\Psi\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\Psi\left(\frac{\mathrm{3}}{\mathrm{2}}+{b}\right)=−\Psi'\left(\frac{\mathrm{3}}{\mathrm{2}}\right){b} \\ $$$$\Leftrightarrow\frac{\mathrm{1}}{\mathrm{12}}\underset{{b}\rightarrow\mathrm{0}} {\mathrm{lim}}{b}\Gamma\left({b}\right)\Psi'\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{12}}\Gamma\left(\mathrm{1}+{b}\right)\Psi'\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\frac{\Psi'\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\mathrm{12}} \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 04/Jul/21

thank you so much..

$${thank}\:{you}\:{so}\:{much}.. \\ $$

Commented by mindispower last updated on 04/Jul/21

pleasur sir

$${pleasur}\:{sir} \\ $$

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