If $I_n=\int_0^{\frac{\pi}{4}} an ^n heta d heta$, then $I_{12}+I_{10}=$

Question

If $I_n=\int_0^{\frac{\pi}{4}} 	an ^n 	heta d 	heta$, then $I_{12}+I_{10}=$

Aicharya · Accepted Answer

$\begin{aligned} & \int_0^{\frac{\pi}{4}}\left(	an ^{\mathrm{n}} x+	an ^{\mathrm{n}-2} xight) \mathrm{d} x=\frac{1}{\mathrm{n}-1} \ & 	herefore \quad \mathrm{I}_{12}+\mathrm{I}_{10}=\int_0^{\frac{\pi}{4}}\left(	an ^{12} 	heta+	an ^{10} 	hetaight) \mathrm{d} 	heta \ &=\frac{1}{12-1} \ &=\frac{1}{11}\end{aligned}$

If $I_n=\int_0^{\frac{\pi}{4}} \tan ^n \theta d \theta$, then $I_{12}+I_{10}=$

Solution