Question Number 37189 by rahul 19 last updated on 10/Jun/18
$$\mathrm{If}\:\int\:\:\frac{\mathrm{cos}\:\mathrm{4}{x}+\mathrm{1}}{\mathrm{cot}\:{x}−\mathrm{tan}\:{x}}\:{dx}={A}\:\mathrm{cos}\:\mathrm{4}{x}+{B},\:\mathrm{then}\: \\ $$$$\mathrm{find}\:\mathrm{A},\mathrm{B}? \\ $$
Answered by math1967 last updated on 10/Jun/18
$$\int\frac{\mathrm{2}{cos}^{\mathrm{2}} \mathrm{2}{x}}{\frac{{cos}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {x}}{{sinxcosx}}}{dx} \\ $$$$\int\frac{\mathrm{2}{cos}^{\mathrm{2}} \mathrm{2}{xsinxcosx}}{{cos}\mathrm{2}{x}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{2}{sin}\mathrm{2}{xcos}\mathrm{2}{xdx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int{sin}\mathrm{4}{xdx}=−\frac{\mathrm{1}}{\mathrm{8}}{cos}\mathrm{4}{x}+{c} \\ $$$$\therefore{A}=−\frac{\mathrm{1}}{\mathrm{8}}\:,{B}={c}={some}\:{constant} \\ $$$${correct}? \\ $$
Commented by MJS last updated on 10/Jun/18
$$\mathrm{correct}! \\ $$