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Question Number 152115 by peter frank last updated on 25/Aug/21

$$\int \frac{\mathrm{dx}}{\mathrm{asin}\mathrm{x}+\mathrm{bcos}\mathrm{x}}$$

Answered by aleks041103 last updated on 26/Aug/21

$$\int \frac{\mathrm{dx}}{\mathrm{asin}\mathrm{x}+\mathrm{bcos}\mathrm{x}}=$$$$=\frac{1}{\sqrt{a^2+b^2}}\int \frac{dx}{sin(x+r)},r=atan\left(b/a\right)$$$$\int \frac{dx}{sinx}=\int \frac{sinxdx}{1-{cos}^{2}x}=-\int \frac{du}{1-u^2}=$$$$=-\frac{1}{2}\int (\frac{1}{1-u}+\frac{1}{1+u})du$$$$=-\frac{1}{2}ln\left(\frac{1+u}{1-u}\right)=ln\sqrt{\frac{1-cosx}{1+cosx}}+C$$$$\int \frac{dx}{asin\left(x\right)+bcos\left(x\right)}=\frac{1}{2\sqrt{a^2+b^2}}ln\left(\frac{1-cos(x+atan\left(b/a\right))}{1+cos(x+atan\left(b/a\right))}\right)+C$$$$$$

Commented by peter frank last updated on 26/Aug/21

$$\mathrm{thank}\mathrm{you}$$

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