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

$${\int }_0^{\frac{\pi }{2}}\frac{dx}{1+\tan {}^{13}x}=?$$

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

$$I={\int }_0^{\frac{\pi }{2}}\frac{{cos}^{13}x}{{cos}^{13}x+{sin}^{13}x}dx$$$$={\int }_0^{\frac{\pi }{2}}\frac{{cos}^{13}(\frac{\pi }{2}-x)}{{cos}^{13}(\frac{\pi }{2}-x)+{sin}^{13}(\frac{\pi }{2}-x)}dx$$$$={\int }_0^{\frac{\pi }{2}}\frac{{sin}^{13}x}{{sin}^{13}x+{cos}^{13}x}dx$$$$2I={\int }_0^{\frac{\pi }{2}}dx$$$$I=\frac{\pi }{4}$$

Commented by rahul 19 last updated on 07/Oct/18

thanks sir ��

Commented by malwaan last updated on 14/Oct/18

$$\mathrm{great}$$

Answered by math1967 last updated on 07/Oct/18

$$I=\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{dx}{1+{tan}^{13}(\frac{\pi }{2}-x)}=\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{dx}{1+{cot}^{13}x}$$$$=\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{dx}{1+\frac{1}{{tan}^{13}x}}=\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{{tan}^{13}x}{1+{tan}^{13}x}dx$$$$\therefore 2I=\underset{0}{\stackrel{\frac{\pi }{2}}{\int }}\frac{1+{tan}^{13}x}{1+{tan}^{13}x}dx={\left[x\right]}_0^{\frac{\pi }{2}}=\frac{\pi }{2}$$$$I=\frac{\pi }{4}ans.$$

Commented by rahul 19 last updated on 07/Oct/18

thanks sir ��

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