Question 209 of 416MediumMCQ
If ∫cos2θ(tan2θ+sec2θ)dθ=λtanθ+2loge∣f(θ)∣+C where C is a constant of integration, then the ordered pair (λ,f(θ)) is equal to:
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Official solution
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 f(\[\theta \]) = 1 + tan\[\theta \]
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