Question Number 22923 by selestian last updated on 24/Oct/17
Answered by $@ty@m last updated on 24/Oct/17
![I_n =∫_0 ^(π/4) tan^n xsec^2 xdx I_n =[((tan^(n+1) x)/(n+1))]_0 ^(π/4) I_1 =[((tan^2 x)/2)]_0 ^(π/4) =(1/2) I_2 =[((tan^3 x)/3)]_0 ^(π/4) =(1/3) I_3 =[((tan^4 x)/4)]_0 ^(π/4) =(1/4) ..... ..... ∴I_1 , I_2 , I_(3,) ....=(1/2), (1/3),(1/4),.... which is an H.P.](https://www.tinkutara.com/question/Q22926.png)
$${I}_{{n}} =\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\mathrm{tan}\:^{{n}} {x}\mathrm{sec}\:^{\mathrm{2}} {xdx} \\ $$$${I}_{{n}} =\left[\frac{\mathrm{tan}\:^{{n}+\mathrm{1}} {x}}{{n}+\mathrm{1}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\right]}} \\ $$$${I}_{\mathrm{1}} =\left[\frac{\mathrm{tan}\:^{\mathrm{2}} {x}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\right]}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =\left[\frac{\mathrm{tan}\:^{\mathrm{3}} {x}}{\mathrm{3}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\right]}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${I}_{\mathrm{3}} =\left[\frac{\mathrm{tan}\:^{\mathrm{4}} {x}}{\mathrm{4}}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\right]}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$….. \\ $$$$….. \\ $$$$\therefore{I}_{\mathrm{1}} ,\:{I}_{\mathrm{2}} ,\:{I}_{\mathrm{3},} ….=\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{1}}{\mathrm{4}},…. \\ $$$${which}\:{is}\:{an}\:{H}.{P}. \\ $$
Commented by selestian last updated on 24/Oct/17
$${tysm}\:{sir}\:{if}\:{you}\:{can}\:{solve}\:{matrix} \\ $$$${Qn}\:{also}\:{then}\:{plz}\:{solve}\:{the}\:{one}\:{i}\:{had} \\ $$$${posted}\:{above} \\ $$$$ \\ $$$$ \\ $$