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Question Number 109872 by bemath last updated on 26/Aug/20

$$\frac{★be★}{\mathcal{M}ath}$$$$\int \frac{\cos xdx}{\sin {}^{2}x+4\sin x-5}?$$

Commented by bemath last updated on 26/Aug/20

$$santuyyallmaster$$

Answered by Sarah85 last updated on 26/Aug/20

$$t=\sin x$$$$\int \frac{dt}{t^2+4t-5}=\frac{1}{6}\ln \frac{t-1}{t+5}...$$

Answered by 1549442205PVT last updated on 26/Aug/20

$$\mathrm{Put}\sin \mathrm{x}=\mathrm{u}\Rightarrow \mathrm{du}=\mathrm{coxdxdx}$$$$\mathrm{F}=\int \frac{\cos xdx}{\sin {}^{2}x+4\sin x-5}=\int \frac{\mathrm{du}}{{\mathrm{u}}^2+4\mathrm{u}-5}$$$$=\int \frac{\mathrm{du}}{(\mathrm{u}-1)(\mathrm{u}+5)}=\frac{1}{6}\int (\frac{1}{\mathrm{u}-1}-\frac{1}{\mathrm{u}+5})\mathrm{du}$$$$=\frac{1}{6}(\ln \mid \mathrm{u}-1\mid -\ln \mid \mathrm{u}+5\mid )=\frac{1}{6}\ln \mid \frac{\mathrm{u}-1}{\mathrm{u}+5}\mid +\mathrm{C}$$$$\mathbf{F}=\frac{1}{6}\mathbf{ln}\mid \frac{\mathbf{sinx}-1}{\mathbf{sinx}+5}\mid +\mathbf{C}$$

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