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Integral Calculus

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13 Feb 2012 23:50:48 IST

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integral calculus solution

Mathematics

what is the integral of square root of cotx dx?

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Vishal U. Karve

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Joined: 18 Jan 2012

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14 Feb 2012 18:34:55 IST

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put cot x=t².

-cosec²x dx = 2 t dt

-(1+t⁴)=2t dt

The integral becomes

Write -2t² as (1-t²) - (1+t²)

You get 2 integrals.

Divide by t² in both numerator and denominator. The 2 integrals become

(Ignore the box in the second integral)

Now in the first integral put t + 1/t =u.

In second integral put t - 1/t = v.

Hope you can do the rest..

jagdish singh

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14 Feb 2012 19:43:50 IST

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$\hspace{-16}$Let $\mathbf{I=\int\sqrt{\tan x}dx}$\\\\\\ and $\mathbf{J=\int\sqrt{\cot x}dx}$\\\\\\ Now $\mathbf{I+J=\int\left(\sqrt{\tan x}+\sqrt{\cot x}\right)dx}$\\\\\\ $\mathbf{=\int\frac{\sin x+\cos x}{\sqrt{\sin x.\cos x}}dx=\sqrt{2}.\int \frac{\left(\sin x+\cos x\right)}{\sqrt{\sin 2x}}dx}$\\\\\\ Now Let $\mathbf{\left(\sin x-\cos x\right)=t\Leftrightarrow \left(\cos x+\sin x\right)dx=dt}$\\\\\\ $\mathbf{=\sqrt{2}\int\frac{1}{\sqrt{1-t^2}}dt=\sqrt{2}.\sin^{-1}\left(t\right)}$\\\\\\ So $\mathbf{I+J=\sqrt{2}.\sin^{-1}\left(\sin x-\cos x\right)}$\\\\\\ Similarly $\mathbf{I-J=\int\left(\sqrt{\tan x}-\sqrt{\cot x}\right)dx}$\\\\\\ $\mathbf{=\int\frac{\sin x-\cos x}{\sqrt{\sin x.\cos x}}dx=\sqrt{2}.\int \frac{\left(\sin x-\cos x\right)}{\sqrt{\sin 2x}}dx}$\\\\\\ Now Let $\mathbf{\left(\sin x+\cos x\right)=u\Leftrightarrow \left(\sin x-\cos x\right)dx=-du}$\\\\\\ $\mathbf{=-\sqrt{2}\int\frac{1}{\sqrt{u^-1}}du=-\sqrt{2}.\ln \left|u+\sqrt{u^2-1}\right|}$\\\\\\$

$\hspace{-16}$So $\mathbf{I-J=-\sqrt{2}.\ln \left|(\sin+\cos x)+\sqrt{\sin 2x}\right|}$\\\\\\ So $\mathbf{I=\int\sqrt{\tan x}dx = \frac{\sqrt{2}}{2}\left\{.\sin^{-1}(\sin x-\cos x)-\ln \left|(\sin+\cos x)+\sqrt{\sin 2x}\right|}+C\right\}$\\\\\\ and $\mathbf{J=\int\sqrt{\cot x}dx = \frac{\sqrt{2}}{2}\left\{.\sin^{-1}(\sin x-\cos x)+\ln \left|(\sin+\cos x)+\sqrt{\sin 2x}\right|}+C\right\}$$

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