QJEEM21Mad975: JEE Main 2021 Mathematics - Class 12 JEEM21Mad975 Solution: If , when C is constant of integration, then the value of is…

Question

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If

\int {{{\sin x} \over {{{\sin }^3}x + {{\cos }^3}x}}dx = }

\alpha {\log _e}|1 + \tan x| + \beta {\log _e}|1 - \tan x + {\tan ^2}x| + \gamma {\tan ^{ - 1}}\left( {{{2\tan x - 1} \over {\sqrt 3 }}} \right) + C

, when C is constant of integration, then the value of

18(\alpha + \beta + {\gamma ^2})

is ______________.

Accepted Answer

= \int {{{{{\sin x} \over {{{\cos }^3}x}}} \over {1 + {{\tan }^3}x}}dx = \int {{{\tan x.{{\sec }^2}x} \over {(\tan x + 1)(1 + {{\tan }^2}x - \tan x)}}dx} }

Let

\tan x = t \Rightarrow {\sec ^2}x.\,dx = dt

= \int {{t \over {(t + 1)({t^2} - t + 1)}}dt}

= \int {\left( {{A \over {t + 1}} + {{B(2t - 1)} \over {{t^2} - t + 1}} + {C \over {{t^2} - t + 1}}} \right)dx}

\Rightarrow A({t^2} - t + 1) + B(2t - 1)({t^2} - t + 1) + C(t + 1) = t

\Rightarrow {t^2}(A + 2B) + t( - A + B + C) + A - B + C = 1

\therefore

A + 2B = 0

..... (1)

- A + B + C = 1

... (2)

A - B + C = 0

... (3)

\Rightarrow C = {1 \over 2} \Rightarrow A - B = - {1 \over 2}

... (4)

A + 2B = 0

A - B = - {1 \over 2}

\Rightarrow

3B = {1 \over 2} \Rightarrow B = {1 \over 6}

A = - {1 \over 3}

I = - {1 \over 3}\int {{{dt} \over {1 + t}} + {1 \over 6}\int {{{2t - 1} \over {{t^2} - t + 1}}dt + {1 \over 2}\int {{{dt} \over {{t^2} - t + 1}}} } }

= - {1 \over 3}\ln |(1 + \tan x)| + {1 \over 6}\ln |{\tan ^2}x - \tan x + 1| + {1 \over 2}.{2 \over {\sqrt 3 }}{\tan ^{ - 1}}\left( {{{\left( {\tan x - {1 \over 2}} \right)} \over {{{\sqrt 3 } \over 2}}}} \right)

= - {1 \over 3}\ln |(1 + \tan x)| + {1 \over 6}\ln |{\tan ^2}x - \tan x + 1| + {1 \over {\sqrt 3 }}{\tan ^{ - 1}}\left( {{{2\tan x - 1} \over {\sqrt 3 }}} \right) + C

\alpha = - {1 \over 3},\beta = {1 \over 6},\gamma = {1 \over {\sqrt 3 }}

18(\alpha + \beta + {\gamma ^2}) = 18\left( { - {1 \over 3} + {1 \over 6} + {1 \over 3}} \right) = 3

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