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Question Number 147593 by 0731619 last updated on 22/Jul/21

Answered by gsk2684 last updated on 22/Jul/21

$$put\sin y=t\Rightarrow \cos ydy=dt$$$$\int \frac{1}{t^2+t-6}dt=\int \frac{1}{(t-2)(t+3)}dt$$$$=\int (\frac{\frac{1}{5}}{t-2}+\frac{-\frac{1}{5}}{t+3})dt$$$$=\frac{1}{5}(\ln (t-2)-\ln (t+3))+c$$$$=\frac{1}{5}\ln \left(\frac{t-2}{t+3}\right)+c$$$$=\frac{1}{5}\ln \left(\frac{\sin y-2}{\sin y+3}\right)+c$$

Commented by 0731619 last updated on 22/Jul/21

$$thanks$$

Answered by Olaf_Thorendsen last updated on 22/Jul/21

$$$$$$f\left(y\right)=\frac{\cos y}{{\sin }^{2}y+\sin y-6}$$$$f\left(y\right)=\frac{\cos y}{(\sin y-2)(\sin y+3)}$$$$f\left(y\right)=\frac{\frac{\cos y}{{(\sin y+3)}^2}}{\left(\frac{\sin y-2}{\sin y+3}\right)}$$$$f\left(y\right)=\frac{1}{5}.\frac{\frac{\cos y(\sin y+3)-(\sin y-2)\cos y}{{(\sin y+3)}^2}}{\left(\frac{\sin y-2}{\sin y+3}\right)}$$$$f\left(y\right)=\frac{1}{5}.\frac{\frac{d}{dy}\left(\frac{\sin y-2}{\sin y+3}\right)}{\left(\frac{\sin y-2}{\sin y+3}\right)}$$$$\Rightarrow \mathrm{F}\left(y\right)=\int f\left(y\right)dy=\frac{1}{5}\ln \mid \frac{\sin y-2}{\sin y+3}\mid +\mathrm{C}$$

Commented by 0731619 last updated on 22/Jul/21

$$thanks$$$$$$

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