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Question Number 45021 by rahul 19 last updated on 07/Oct/18

∫_0 ^(π/2) (dx/(1+tan^(13) x)) = ?

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{13}} {x}}\:=\:? \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Oct/18

I=∫_0 ^(π/2) ((cos^(13) x)/(cos^(13) x+sin^(13) x))dx =∫_0 ^(π/2) ((cos^(13) ((π/2)−x))/(cos^(13) ((π/2)−x)+sin^(13) ((π/2)−x)))dx =∫_0 ^(π/2) ((sin^(13) x)/(sin^(13) x+cos^(13) x))dx 2I=∫_0 ^(π/2) dx I=(π/4)

$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}^{\mathrm{13}} {x}}{{cos}^{\mathrm{13}} {x}+{sin}^{\mathrm{13}} {x}}{dx} \\ $$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}^{\mathrm{13}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)}{{cos}^{\mathrm{13}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)+{sin}^{\mathrm{13}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)}{dx} \\ $$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}^{\mathrm{13}} {x}}{{sin}^{\mathrm{13}} {x}+{cos}^{\mathrm{13}} {x}}{dx} \\ $$ $$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx} \\ $$ $${I}=\frac{\pi}{\mathrm{4}} \\ $$

Commented byrahul 19 last updated on 07/Oct/18

thanks sir ��

Commented bymalwaan last updated on 14/Oct/18

great

$$\mathrm{great} \\ $$

Answered by math1967 last updated on 07/Oct/18

I=∫_0 ^(π/2) (dx/(1+tan^(13) ((π/2)−x)))=∫_0 ^(π/2) (dx/(1+cot^(13) x)) =∫_0 ^(π/2) (dx/(1+(1/(tan^(13) x))))=∫_0 ^(π/2) ((tan^(13) x)/(1+tan^(13) x))dx ∴2I=∫_0 ^(π/2) ((1+tan^(13) x)/(1+tan^(13) x))dx=[x]_(0 ) ^(π/2) =(π/2) I=(π/4) ans.

$${I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{dx}}{\mathrm{1}+{tan}^{\mathrm{13}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{dx}}{\mathrm{1}+{cot}^{\mathrm{13}} {x}} \\ $$ $$=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{dx}}{\mathrm{1}+\frac{\mathrm{1}}{{tan}^{\mathrm{13}} {x}}}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{{tan}^{\mathrm{13}} {x}}{\mathrm{1}+{tan}^{\mathrm{13}} {x}}{dx} \\ $$ $$\therefore\mathrm{2}{I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{1}+{tan}^{\mathrm{13}} {x}}{\mathrm{1}+{tan}^{\mathrm{13}} {x}}{dx}=\left[{x}\right]_{\mathrm{0}\:\:} ^{\frac{\pi}{\mathrm{2}}} =\frac{\pi}{\mathrm{2}} \\ $$ $${I}=\frac{\pi}{\mathrm{4}}\:{ans}. \\ $$

Commented byrahul 19 last updated on 07/Oct/18

thanks sir ��

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