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Integratesin x dx /sin 5x&1 dx / (sin5x+cos5x)

Shraddhey Srivastava , 15 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:

$I = \int \frac{sinx}{sin5x}dx$

$I = \int sinx.cosec5x.dx$

Simplify it using trigonometry identity, we have

$I = \int \frac{sec^{4}x}{2+2sec^{2}x+sec^{4}x-12tan^{2}x-2sec^{2}x.tan^{2}x+2tan^{4}x}dx$ $tanx = t$

$sec^{2}x.dx = dt$

$I = \int \frac{t^{2}+1}{t^{4}-10t^{2}+5}dt$

$I = \int (\frac{\sqrt{5}-3}{2\sqrt{5}(t^{2}+2\sqrt{5}-5)}+\frac{-3-\sqrt{5}}{2\sqrt{5}(-t^{2}+2\sqrt{5}+5})dt$

$I =\frac{ (\sqrt{5+\sqrt{5}})tanh^{-1}(\frac{(\sqrt{5}-3)t}{\sqrt{10-2\sqrt{5}}})+(\sqrt{5-\sqrt{5}})tanh^{-1}(\frac{(3+\sqrt{5})t}{\sqrt{10+2\sqrt{5}}})}{5\sqrt{2}}+cons$

$I =\frac{ (\sqrt{5+\sqrt{5}})tanh^{-1}(\frac{(\sqrt{5}-3)tanx}{\sqrt{10-2\sqrt{5}}})+(\sqrt{5-\sqrt{5}})tanh^{-1}(\frac{(3+\sqrt{5})tanx}{\sqrt{10+2\sqrt{5}}})}{5\sqrt{2}}+cons$