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Integral Calculus> what is the integral of(1-tan 2 x) 1/2 dx???

question mark

what is the integral of(1-tan2x)1/2dx???

Mostafijur Rahaman , 10 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:

$I = \int \sqrt{1-tan^{2}x}dx$

$t = tanx$

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

$dx = \frac{dt}{1+t^{2}}$

$I = \int \frac{\sqrt{1-t^{2}}}{1+t^{2}}dt$

$t = sin(s)$

$dt = cos(s).ds$

$I = \int \frac{cos^{2}(s)}{1+sin^{2}(s)}ds$

$I = \int \frac{sec^{2}(s)}{sec^{4}(s)+sec^{2}(s).tan^{2}(s)}ds$

$u = tan(s)$

$du = sec^{2}(s).ds$

$I = \int \frac{1}{2u^{4}+3u^{2}+1}du$

Using the partial fraction rule, we have

$I = \sqrt{2}tan^{-1}(\sqrt{2}u)-tan^{-1}(u)+constant$

$I = \sqrt{2}tan^{-1}(\sqrt{2}tan(s))-s+constant$

$I = \sqrt{2}tan^{-1}(\frac{\sqrt{2}t}{\sqrt{1-t^{2}}})-sin^{-1}(t)+constant$

$I = \sqrt{2}tan^{-1}(\frac{\sqrt{2}tan(x)}{\sqrt{1-tan^{2}(x)}})-sin^{-1}(tan(x))+constant$

Thanks & Regards

Jitender Singh

IIT Delhi

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