Let $y=\sin \phi$ $\Rightarrow d y=\cos \phi d \phi$ Therefore, $\int \frac{(3 \sin \phi-2) \cos \phi}{5-\cos ^{2} \phi-4 \sin \phi} d \phi$ = $\int \frac{(3 y-2) d y}{5-\left(1-y^{2}\right)-4 y}$ = $\int \frac{3 y-2}{y^{2}-4 y+4} d y$ Now, we write $\frac{3 y-2}{(y-2)^{2}}=\frac{A}{y-2}+\frac{B}{(y-2)^{2}}$ Therefore, 3y - 2 = A (y - 2) + B Comparing the coefficients of y and constant term, we get A = 3 and B = 4. for y = 2, y= 0 Therefore, the required integral is given by $I=\int\left[\frac{3}{y-2}+\frac{4}{(y-2)^{2}}\right] d y$ = $3 \int \frac{d y}{y-2}+4 \int \frac{d y}{(y-2)^{2}}$ = $3 \log |y-2|+4\left(-\frac{1}{y-2}\right)+C$ = $3 \log |\sin \phi-2|+\frac{4}{2-\sin \phi}+C$ = $3 \log (2-\sin \phi)+\frac{4}{2-\sin \phi}+C$ (since, sin $\phi $ $\in$[-1,1], sin $\phi $< 2, 2 - sin $\phi $ is always positive)
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.