# 1-sin-3-x-sin-a-x-

Question Number 76929 by peter frank last updated on 01/Jan/20
$$\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{sin}\:^{\mathrm{3}} {x}\left(\mathrm{sin}\left({a}+{x}\right)\right)\:}} \\$$
Answered by MJS last updated on 01/Jan/20
$$\int\frac{{dx}}{\:\sqrt{\mathrm{sin}^{\mathrm{3}} \:{x}\:\mathrm{sin}\:\left({a}+{x}\right)}}= \\$$$$=\int\frac{{dx}}{\:\sqrt{\mathrm{sin}^{\mathrm{3}} \:{x}\:\left(\mathrm{sin}\:{a}\:\mathrm{cos}\:{x}\:+\mathrm{cos}\:{a}\:\mathrm{sin}\:{x}\right)}}= \\$$$$\:\:\:\:\:\left[\mathrm{sin}\:{a}\:={q};\:\mathrm{cos}\:{a}\:={p}\right] \\$$$$=\int\frac{{dx}}{\:\sqrt{{p}\mathrm{sin}^{\mathrm{4}} \:{x}\:+{q}\:\mathrm{cos}\:{x}\:\mathrm{sin}^{\mathrm{3}} \:{x}}}= \\$$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:\rightarrow\:{dx}=\mathrm{cos}^{\mathrm{2}} \:{x}\:{dt}=\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\$$$$=\int\frac{{dt}}{\:\sqrt{{t}^{\mathrm{3}} \left({pt}+{q}\right)}}= \\$$$$\:\:\:\:\:\left[{u}=\sqrt{{pt}+{q}}\:\rightarrow\:{dt}=\frac{\mathrm{2}\sqrt{{pt}+{q}}}{{p}}\right] \\$$$$=\mathrm{2}\sqrt{{p}}\int\frac{{du}}{\left({u}^{\mathrm{2}} −{q}\right)^{\mathrm{3}/\mathrm{2}} }=−\frac{\mathrm{2}\sqrt{{p}}}{{q}}×\frac{{u}}{\:\sqrt{{u}^{\mathrm{2}} −{p}}}= \\$$$$=−\frac{\mathrm{2}}{{q}}\sqrt{{p}+\frac{{q}}{{t}}}=−\frac{\mathrm{2}}{{q}}\sqrt{{p}+{q}\mathrm{cot}\:{x}}= \\$$$$=−\frac{\mathrm{2}}{\mathrm{sin}\:{a}}\sqrt{\mathrm{cos}\:{a}\:+\mathrm{sin}\:{a}\:\mathrm{cot}\:{x}}+{C} \\$$
Commented by peter frank last updated on 01/Jan/20
$${thank}\:{you} \\$$
Answered by petrochengula last updated on 03/Jan/20
$$=\int\frac{\mathrm{1}}{\:\sqrt{\frac{{sin}^{\mathrm{4}} {x}\left({sinacosx}+{sinxcosa}\right.}{{sinx}}}} \\$$$$=\int\frac{\mathrm{1}}{{sin}^{\mathrm{2}} {x}\sqrt{{sinacotx}+{cosa}}}{dx} \\$$$$=\int\frac{{cosec}^{\mathrm{2}} {x}}{\:\sqrt{{sinacotx}+{cosa}}}{dx} \\$$$$=−\int\frac{−{cosec}^{\mathrm{2}} {x}}{\:\sqrt{{sinacotx}+{cosa}}}{dx} \\$$$${let}\:{t}=\sqrt{{sinacotx}+{cosa}} \\$$$${game}\:{over} \\$$