( 3 - 2 ) 5 - ^2 - 4 , d is equal to - Vidyadip

MCQ

$\int {\frac{{\left( {3\sin \phi - 2} \right)\cos \phi }}{{5 - {{\cos }^2}\phi - 4\sin \phi }}\,} d\phi$ is equal to

A
$3\log \left( {2 - \sin \phi } \right)\frac{4}{{\left( {\sin \phi - 2} \right)}} + C$
B
$3\log \left( {\sin \phi - 2} \right) + \frac{4}{{\left( {2 - \sin \phi } \right)}} + C$
C
$\log \left( {2 - \sin \phi } \right) + \frac{4}{{\left( {2 - \sin \phi } \right)}} + C$
✓
$3\log \left( {2 - \sin \phi } \right) + \frac{4}{{\left( {2 - \sin \phi } \right)}} + C$

✓

Answer

Correct option: D.

$3\log \left( {2 - \sin \phi } \right) + \frac{4}{{\left( {2 - \sin \phi } \right)}} + C$

d
Put $\sin \phi=t ;$ then $\cos \phi \,d\phi = dt$

$I=\int \frac{(3 t-2) d t}{5-\left(1-t^{2}\right)-4 t}$

$I=\int \frac{3 t-2}{t^{2}-4 t+4} d t=\int \frac{3 t-2}{\left(t^{2}-2\right)^{2}} d t$

$\frac{3 t-2}{(t-2)^{2}}=\frac{A}{(t-2)}+\frac{B}{(t-2)^{2}}$

$(3 t-2)=A(t-2)+B$

$\therefore A=3, B=4$

$ \therefore \mathrm{I}=\int \frac{3}{\mathrm{t}-2} \mathrm{dt}+\int \frac{4}{(\mathrm{t}-2)^{2}} \mathrm{dt}$

$=3 \log |(\sin \phi-2)|+\frac{-4}{(\mathrm{t}-2)}+\mathrm{C}$

$=3 \log (2-\sin \phi)+\frac{4}{(2-\sin \phi)}+\mathrm{C} $

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