If ()/(5+7-2^2),d=A_e|B()|+C, where C is a constant of integration, then B()A can be:

Question

If $\displaystyle\int\frac{\cos	heta}{5+7\sin	heta-2\cos^2	heta}\,d	heta=A\log_e|B(	heta)|+C$, where C is a constant of integration, then $\dfrac{B(	heta)}A$ can be:

Accepted Answer

$$\int {{{\cos \theta } \over {5 + 7\sin \theta  - 2{{\cos }^2}\theta }}} d\theta $$

= $$\int {{{\cos \theta d\theta } \over {5 + 7\sin \theta  - 2\left( {1 - {{\sin }^2}\theta } \right)}}} $$

= $$\int {{{\cos \theta d\theta } \over {3 + 7\sin \theta  + 2{{\sin }^2}\theta }}} $$

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($$\theta $$) = $${{{2\sin \theta  + 1} \over {\sin \theta  + 3}}}$$

$$ \therefore $$ $${{{B\left( \theta  \right)} \over A}}$$ = $${{5\left( {2\sin \theta  + 1} \right)} \over {\sin \theta  + 3}}$$

Solutions · 1