JEE MAIN - Mathematics (2020 - 9th January Evening Slot - No. 6)
If $$\int {{{d\theta } \over {{{\cos }^2}\theta \left( {\tan 2\theta + \sec 2\theta } \right)}}} = \lambda \tan \theta + 2{\log _e}\left| {f\left( \theta \right)} \right| + C$$
where C is a constant of integration, then the ordered pair ($$\lambda $$, ƒ($$\theta $$)) is equal to :
where C is a constant of integration, then the ordered pair ($$\lambda $$, ƒ($$\theta $$)) is equal to :
(–1, 1 – tan$$\theta $$)
(1, 1 + tan$$\theta $$)
(–1, 1 + tan$$\theta $$)
(1, 1 – tan$$\theta $$)
Explanation
I = $$\int {{{d\theta } \over {{{\cos }^2}\theta \left( {\tan 2\theta + \sec 2\theta } \right)}}}$$
= $$\int {{{{{\sec }^2}\theta d\theta } \over {{{2\tan \theta } \over {1 - {{\tan }^2}\theta }} + {{1 + {{\tan }^2}\theta } \over {1 - {{\tan }^2}\theta }}}}} $$
= $$\int {{{\left( {1 - {{\tan }^2}\theta } \right){{\sec }^2}\theta d\theta } \over {{{\left( {1 + {{\tan }}\theta } \right)}^2}}}} $$
tan$$\theta $$ = t $$ \Rightarrow $$ sec2 $$\theta $$ d$$\theta $$ = dt
= $$\int {{{\left( {1 - {t^2}} \right)dt} \over {{{\left( {1 + {t}} \right)}^2}}}} $$
= $$\int {{{\left( {1 - t} \right)\left( {1 + t} \right)dt} \over {{{\left( {1 + {t}} \right)}^2}}}} $$
= $$\int {{{\left( {1 - t} \right)} \over {\left( {1 + t} \right)}}} dt$$
= $$\int {\left( {{1 \over {\left( {1 + t} \right)}} - {t \over {\left( {1 + t} \right)}}} \right)} dt$$
= $${\log _e}\left| {1 + t} \right|$$ - $$\int {\left( {{{1 + t} \over {\left( {1 + t} \right)}} - {1 \over {\left( {1 + t} \right)}}} \right)} dt$$
= $${\log _e}\left| {1 + t} \right|$$ - t + $${\log _e}\left| {1 + t} \right|$$ + C
= 2$${\log _e}\left| {1 + t} \right|$$ - t + C
= 2$${\log _e}\left| {1 + \tan \theta } \right|$$ - tan $$\theta $$ + C
$$ \therefore $$ $$\lambda $$ = -1 and ƒ($$\theta $$) = 1 + tan$$\theta $$
= $$\int {{{{{\sec }^2}\theta d\theta } \over {{{2\tan \theta } \over {1 - {{\tan }^2}\theta }} + {{1 + {{\tan }^2}\theta } \over {1 - {{\tan }^2}\theta }}}}} $$
= $$\int {{{\left( {1 - {{\tan }^2}\theta } \right){{\sec }^2}\theta d\theta } \over {{{\left( {1 + {{\tan }}\theta } \right)}^2}}}} $$
tan$$\theta $$ = t $$ \Rightarrow $$ sec2 $$\theta $$ d$$\theta $$ = dt
= $$\int {{{\left( {1 - {t^2}} \right)dt} \over {{{\left( {1 + {t}} \right)}^2}}}} $$
= $$\int {{{\left( {1 - t} \right)\left( {1 + t} \right)dt} \over {{{\left( {1 + {t}} \right)}^2}}}} $$
= $$\int {{{\left( {1 - t} \right)} \over {\left( {1 + t} \right)}}} dt$$
= $$\int {\left( {{1 \over {\left( {1 + t} \right)}} - {t \over {\left( {1 + t} \right)}}} \right)} dt$$
= $${\log _e}\left| {1 + t} \right|$$ - $$\int {\left( {{{1 + t} \over {\left( {1 + t} \right)}} - {1 \over {\left( {1 + t} \right)}}} \right)} dt$$
= $${\log _e}\left| {1 + t} \right|$$ - t + $${\log _e}\left| {1 + t} \right|$$ + C
= 2$${\log _e}\left| {1 + t} \right|$$ - t + C
= 2$${\log _e}\left| {1 + \tan \theta } \right|$$ - tan $$\theta $$ + C
$$ \therefore $$ $$\lambda $$ = -1 and ƒ($$\theta $$) = 1 + tan$$\theta $$
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