If \int {{{d heta } \over {{{\cos }^2} heta \left( { an 2 heta + \sec 2 heta } ight)}}} = \lambda an heta + 2{\log _e}\left| {f\left( heta ight)} ight| + C where C is a constant of integration, then the ordered pair (\lambda , ƒ( heta )) is equal to :

Question

If \int {{{d	heta } \over {{{\cos }^2}	heta \left( {	an 2	heta + \sec 2	heta } ight)}}} = \lambda 	an 	heta + 2{\log _e}\left| {f\left( 	heta ight)} ight| + C where C is a constant of integration, then the ordered pair (\lambda , ƒ(	heta )) is equal to :

Accepted Answer

Answer: C. (–1, 1 + tan	heta ) I = \int {{{d	heta } \over {{{\cos }^2}	heta \left( {	an 2	heta + \sec 2	heta } ight)}}} = \int {{{{{\sec }^2}	heta d	heta } \over {{{2	an 	heta } \over {1 - {{	an }^2}	heta }} + {{1 + {{	an }^2}	heta } \over {1 - {{	an }^2}	heta }}}}} = \int {{{\left( {1 - {{	an }^2}	heta } ight){{\sec }^2}	heta d	heta } \over {{{\left( {1 + {{	an }}	heta } ight)}^2}}}} tan	heta = t \Rightarrow sec2 	heta d	heta = dt = \int {{{\left( {1 - {t^2}} ight)dt} \over {{{\left( {1 + {t}} ight)}^2}}}} = \int {{{\left( {1 - t} ight)\left( {1 + t} ight)dt} \over {{{\left( {1 + {t}} ight)}^2}}}} = \int {{{\left( {1 - t} ight)} \over {\left( {1 + t} ight)}}} dt = \int {\left( {{1 \over {\left( {1 + t} ight)}} - {t \over {\left( {1 + t} ight)}}} ight)} dt = {\log _e}\left| {1 + t} ight| - \int {\left( {{{1 + t} \over {\left( {1 + t} ight)}} - {1 \over {\left( {1 + t} ight)}}} ight)} dt = {\log _e}\left| {1 + t} ight| - t + {\log _e}\left| {1 + t} ight| + C = 2{\log _e}\left| {1 + t} ight| - t + C = 2{\log _e}\left| {1 + 	an 	heta } ight| - tan 	heta + C 	herefore \lambda = -1 and ƒ(	heta ) = 1 + tan	heta

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 :

Answer

Explanation

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