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				   Integrate
sin x dx /sin 5x
&
1 dx / (sin5x+cos5 x)


6 years ago

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### Answers : (1)

										Ans:$I = \int \frac{sinx}{sin5x}dx$$I = \int sinx.cosec5x.dx$Simplify it using trigonometry identity, we have$I = \int \frac{sec^{4}x}{2+2sec^{2}x+sec^{4}x-12tan^{2}x-2sec^{2}x.tan^{2}x+2tan^{4}x}dx$$tanx = t$$sec^{2}x.dx = dt$$I = \int \frac{t^{2}+1}{t^{4}-10t^{2}+5}dt$$I = \int (\frac{\sqrt{5}-3}{2\sqrt{5}(t^{2}+2\sqrt{5}-5)}+\frac{-3-\sqrt{5}}{2\sqrt{5}(-t^{2}+2\sqrt{5}+5})dt$$I =\frac{ (\sqrt{5+\sqrt{5}})tanh^{-1}(\frac{(\sqrt{5}-3)t}{\sqrt{10-2\sqrt{5}}})+(\sqrt{5-\sqrt{5}})tanh^{-1}(\frac{(3+\sqrt{5})t}{\sqrt{10+2\sqrt{5}}})}{5\sqrt{2}}+cons$$I =\frac{ (\sqrt{5+\sqrt{5}})tanh^{-1}(\frac{(\sqrt{5}-3)tanx}{\sqrt{10-2\sqrt{5}}})+(\sqrt{5-\sqrt{5}})tanh^{-1}(\frac{(3+\sqrt{5})tanx}{\sqrt{10+2\sqrt{5}}})}{5\sqrt{2}}+cons$Second integrand is not clear.Thanks & RegardsJitender SinghIIT DelhiaskIITians Faculty

2 years ago

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