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Question Number 76929 by peter frank last updated on 01/Jan/20

$$\int \frac{1}{\sqrt{\sin {}^{3}x\left(\sin (a+x)\right)}}$$

Answered by MJS last updated on 01/Jan/20

$$\int \frac{dx}{\sqrt{{\sin }^{3}x\sin (a+x)}}=$$$$=\int \frac{dx}{\sqrt{{\sin }^{3}x(\sin a\cos x+\cos a\sin x)}}=$$$$[\sin a=q;\cos a=p]$$$$=\int \frac{dx}{\sqrt{p{\sin }^{4}x+q\cos x{\sin }^{3}x}}=$$$$[t=\tan x\to dx={\cos }^{2}xdt=\frac{dt}{t^2+1}]$$$$=\int \frac{dt}{\sqrt{t^3(pt+q)}}=$$$$[u=\sqrt{pt+q}\to dt=\frac{2\sqrt{pt+q}}{p}]$$$$=2\sqrt{p}\int \frac{du}{{(u^2-q)}^{3/2}}=-\frac{2\sqrt{p}}{q}\times \frac{u}{\sqrt{u^2-p}}=$$$$=-\frac{2}{q}\sqrt{p+\frac{q}{t}}=-\frac{2}{q}\sqrt{p+q\cot x}=$$$$=-\frac{2}{\sin a}\sqrt{\cos a+\sin a\cot x}+C$$

Commented by peter frank last updated on 01/Jan/20

$$thankyou$$

Answered by petrochengula last updated on 03/Jan/20

$$=\int \frac{1}{\sqrt{\frac{{sin}^{4}x(sinacosx+sinxcosa}{sinx}}}$$$$=\int \frac{1}{{sin}^{2}x\sqrt{sinacotx+cosa}}dx$$$$=\int \frac{{cosec}^{2}x}{\sqrt{sinacotx+cosa}}dx$$$$=-\int \frac{-{cosec}^{2}x}{\sqrt{sinacotx+cosa}}dx$$$$lett=\sqrt{sinacotx+cosa}$$$$gameover$$

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