12.0 Reduction Formula

Question 27. Let ${I_n} = \int {{{\left( {\tan x} \right)}^n}dx} $ then prove that $${I_n} + {I_{n - 2}} = {{{{\left( {\tan x} \right)}^{n - 1}}} \over {n - 1}} + C$$

Solution: $${I_{n - 2}} = \int {{{\left( {\tan x} \right)}^{n - 2}}dx} $$ $${I_n} + {I_{n - 2}} = \int {{{\left( {\tan x} \right)}^n} + {{\left( {\tan x} \right)}^{n - 2}}dx} $$ $${I_n} + {I_{n - 2}} = \int {{{\left( {\tan x} \right)}^{n - 2}}\left( {{{\tan }^2}x + 1} \right)dx} $$ $${I_n} + {I_{n - 2}} = \int {{{\left( {\tan x} \right)}^{n - 2}}{{\sec }^2}xdx} $$ Let $\tan x = t$ $${\sec ^2}xdx = dt$$ $${I_n} + {I_{n - 2}} = \int {{{\left( {\mathop{\rm t}\nolimits} \right)}^{n - 2}}dt} $$ $${I_n} + {I_{n - 2}} = {{{{\left( {\tan x} \right)}^{n - 1}}} \over {n - 1}} + C$$

Question 28. Let ${I_n} = \int\limits_0^{{\pi \over 2}} {{{\left( {\sin x} \right)}^n}dx} $ then find ${{{I_{25}}} \over {{I_{23}}}}$

Solution: $${I_{n - 2}} = \int\limits_0^{{\pi \over 2}} {{{\left( {\sin x} \right)}^{n - 2}}dx} $$ $${I_n} = \int\limits_0^{{\pi \over 2}} {\sin x{{\left( {\sin x} \right)}^{n - 1}}dx} $$ $${I_n} = \left[ {{{\left( {\sin x} \right)}^{n - 1}}\int {\sin xdx} } \right]_0^{{\pi \over 2}} - \int\limits_0^{{\pi \over 2}} {\left\{ {\left( {n - 1} \right){{\left( {\sin x} \right)}^{n - 2}}\cos x\int {\sin xdx} } \right\}dx} $$ $${I_n} = \left[ { - {{\left( {\sin x} \right)}^{n - 1}}\cos x} \right]_0^{{\pi \over 2}} + \int\limits_0^{{\pi \over 2}} {\left\{ {\left( {n - 1} \right){{\left( {\sin x} \right)}^{n - 2}}{{\cos }^2}x} \right\}dx} $$ $${I_n} = \left[ {0 - 0} \right] + \left( {n - 1} \right)\int\limits_0^{{\pi \over 2}} {{{\left( {\sin x} \right)}^{n - 2}}\left( {1 - {{\sin }^2}x} \right)dx} $$ $${I_n} = \left( {n - 1} \right)\int\limits_0^{{\pi \over 2}} {{{\left( {\sin x} \right)}^{n - 2}}dx} - \left( {n - 1} \right)\int\limits_0^{{\pi \over 2}} {{{\left( {\sin x} \right)}^n}dx} $$ $$n{I_n} = \left( {n - 1} \right){I_{n - 2}}$$ $${{{I_n}} \over {{I_{n - 2}}}} = {{n - 1} \over n}$$ $${{{I_{25}}} \over {{I_{23}}}} = {{25 - 1} \over {25}} = {{24} \over {25}}$$

Question 29. Let ${I_n} = \int\limits_0^1 {{x^n}{{\tan }^{ - 1}}xdx} $ then prove that $$\left( {n + 1} \right){I_n} + \left( {n - 1} \right){I_{n - 2}} = {\pi \over 2} - {1 \over n}$$

Solution: $${I_n} = \left[ {{{\tan }^{ - 1}}x\int {{x^n}dx} } \right]_0^1 - \int\limits_0^1 {\left\{ {{1 \over {1 + {x^2}}}\int {{x^n}dx} } \right\}dx} $$ $${I_n} = \left[ {{{{x^{n + 1}}} \over {n + 1}}{{\tan }^{ - 1}}x} \right]_0^1 - \int\limits_0^1 {{{{x^{n + 1}}} \over {n + 1}} \cdot {1 \over {1 + {x^2}}}dx} $$ $$\left( {n + 1} \right){I_n} = \left[ {{x^{n + 1}}{{\tan }^{ - 1}}x} \right]_0^1 - \int\limits_0^1 {{{{x^{n + 1}}} \over {1 + {x^2}}}dx} $$ Similarly, $$\left( {n - 1} \right){I_{n - 2}} = \left[ {{x^{n - 1}}{{\tan }^{ - 1}}x} \right]_0^1 - \int\limits_0^1 {{{{x^{n - 1}}} \over {1 + {x^2}}}dx} $$ Adding both we get $$\left( {n + 1} \right){I_n} + \left( {n - 1} \right){I_{n - 2}} = \left[ {{x^{n + 1}}{{\tan }^{ - 1}}x + {x^{n - 1}}{{\tan }^{ - 1}}x} \right]_0^1 - \int\limits_0^1 {{{{x^{n + 1}} + {x^{n - 1}}} \over {1 + {x^2}}}dx} $$ $$\left( {n + 1} \right){I_n} + \left( {n - 1} \right){I_{n - 2}} = \left[ {\left( {1 + {x^2}} \right){x^{n - 1}}{{\tan }^{ - 1}}x} \right]_0^1 - \int\limits_0^1 {{x^{n - 1}}dx} $$ $$\left( {n + 1} \right){I_n} + \left( {n - 1} \right){I_{n - 2}} = \left[ {2{\pi \over 4} - 0} \right] - \left[ {{{{x^n}} \over n}} \right]_0^1$$ $$\left( {n + 1} \right){I_n} + \left( {n - 1} \right){I_{n - 2}} = {\pi \over 2} - {1 \over n}$$