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Integral Calculus

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5 Feb 2012 14:40:47 IST

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indefinite integration

Mathematics

integration of 1/(sinx+secx)

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jagdish singh

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5 Feb 2012 16:05:37 IST

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thamizhselvan

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6 Feb 2012 15:39:19 IST

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indefinite integration

jagdish singh

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6 Feb 2012 17:04:16 IST

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$\hspace{-16}\int\frac{1}{\sin x+\sec x}dx=\frac{1}{2}.\int\frac{2\cos x}{\sin x.\cos x+1}dx$\\\\\\\\ $\frac{1}{2}.\int\frac{\cos x+\sin x}{\sin x.\cos x+1}dx+\frac{1}{2}.\int\frac{\cos x-\sin x}{\sin x.\cos x+1}dx$\\\\\\\\\ Now Let $I_{1}=\frac{1}{2}.\int\frac{\cos x+\sin x}{\sin x.\cos x+1}dx$\\\\\\\\ Let $\sin x-\cos x=t\Leftrightarrow (\cos x+\sin x)dx=dt$\\\\\\\\ $\int \frac{1}{(\sqrt{3})^2-t^2}dt=\frac{1}{2\sqrt{3}}\ln\left|\frac{t-\sqrt{3}}{t+\sqrt{3}}\right|+C_{1}$\\\\\\\\ Similarly $I_{2}=\frac{1}{2}.\int\frac{\cos x-\sin x}{\sin x.\cos x+1}dx$\\\\\\\\ Let $\sin x+\cos x=u\Leftrightarrow (\cos x-\sin x)dx=du$\\\\\\\\ So $I_{2}=\frac{1}{u^2+1}du=\tan^{-1}(u)+C_{2}$\\\\\\\\ $\int\frac{1}{\sin x+\sec x}dx=\frac{1}{2\sqrt{3}}\ln\left|\frac{\sin x-\cos x-\sqrt{3}}{\sin x-\cos x+\sqrt{3}}\right|+\tan^{-1}\left(\sin x+\cos x\right)+C$$

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