JEE MAIN - Mathematics (2020 - 5th September Evening Slot - No. 15)
If
$$\int {{{\cos \theta } \over {5 + 7\sin \theta - 2{{\cos }^2}\theta }}} d\theta $$ = A$${\log _e}\left| {B\left( \theta \right)} \right| + C$$,
where C is a constant of integration, then $${{{B\left( \theta \right)} \over A}}$$
can be :
$$\int {{{\cos \theta } \over {5 + 7\sin \theta - 2{{\cos }^2}\theta }}} d\theta $$ = A$${\log _e}\left| {B\left( \theta \right)} \right| + C$$,
where C is a constant of integration, then $${{{B\left( \theta \right)} \over A}}$$
can be :
$${{2\sin \theta + 1} \over {5\left( {\sin \theta + 3} \right)}}$$
$${{2\sin \theta + 1} \over {\sin \theta + 3}}$$
$${{5\left( {2\sin \theta + 1} \right)} \over {\sin \theta + 3}}$$
$${{5\left( {\sin \theta + 3} \right)} \over {2\sin \theta + 1}}$$
Explanation
$$\int {{{\cos \theta } \over {5 + 7\sin \theta - 2{{\cos }^2}\theta }}} d\theta $$
= $$\int {{{\cos \theta d\theta } \over {5 + 7\sin \theta - 2\left( {1 - {{\sin }^2}\theta } \right)}}} $$
= $$\int {{{\cos \theta d\theta } \over {3 + 7\sin \theta + 2{{\sin }^2}\theta }}} $$
Let sin $$\theta $$ = t
$$ \Rightarrow $$ cos$$\theta $$d$$\theta $$ = dt
= $$\int {{{dt} \over {3 + 7t + 2{t^2}}}} $$
= $$\int {{{dt} \over {\left( {2t + 1} \right)\left( {t + 3} \right)}}} $$
= $${1 \over 5}\int {\left( {{2 \over {2t + 1}} - {1 \over {t + 3}}} \right)dt} $$
= $${1 \over 5}\ln \left| {{{2t + 1} \over {t + 3}}} \right|$$ + C
= $${1 \over 5}\ln \left| {{{2\sin \theta + 1} \over {\sin \theta + 3}}} \right|$$ + C
$$ \therefore $$ A = $${{1 \over 5}}$$ and B($$\theta $$) = $${{{2\sin \theta + 1} \over {\sin \theta + 3}}}$$
$$ \therefore $$ $${{{B\left( \theta \right)} \over A}}$$ = $${{5\left( {2\sin \theta + 1} \right)} \over {\sin \theta + 3}}$$
= $$\int {{{\cos \theta d\theta } \over {5 + 7\sin \theta - 2\left( {1 - {{\sin }^2}\theta } \right)}}} $$
= $$\int {{{\cos \theta d\theta } \over {3 + 7\sin \theta + 2{{\sin }^2}\theta }}} $$
Let sin $$\theta $$ = t
$$ \Rightarrow $$ cos$$\theta $$d$$\theta $$ = dt
= $$\int {{{dt} \over {3 + 7t + 2{t^2}}}} $$
= $$\int {{{dt} \over {\left( {2t + 1} \right)\left( {t + 3} \right)}}} $$
= $${1 \over 5}\int {\left( {{2 \over {2t + 1}} - {1 \over {t + 3}}} \right)dt} $$
= $${1 \over 5}\ln \left| {{{2t + 1} \over {t + 3}}} \right|$$ + C
= $${1 \over 5}\ln \left| {{{2\sin \theta + 1} \over {\sin \theta + 3}}} \right|$$ + C
$$ \therefore $$ A = $${{1 \over 5}}$$ and B($$\theta $$) = $${{{2\sin \theta + 1} \over {\sin \theta + 3}}}$$
$$ \therefore $$ $${{{B\left( \theta \right)} \over A}}$$ = $${{5\left( {2\sin \theta + 1} \right)} \over {\sin \theta + 3}}$$
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