Evaluate$\int \sqrt{\tan x}dx$
Answer
Let $$A=\int\sqrt{\tan x}\,\mathrm dx$$ $$B=\int\sqrt{\cot x}\,\mathrm dx$$
Then
\[A+B=\int\left(\sqrt{\tan x}+\sqrt{\cot x}\right) dx=\sqrt{2} \int\frac{\sin x+\cos x}{\sqrt{\sin2x}}\mathrm dx\]
\[ =\sqrt2\int\frac{(\sin x-\cos x)'}{\sqrt{1-(\sin x-\cos x)^2}}\,\mathrm dx =\sqrt2\int\frac{1}{\sqrt{1-u^2}}\,\mathrm du\]
\[ =\sqrt2\sin^{-1}u =\sqrt2\sin^{-1}(\sin x-\cos x).\]
Also
\[A-B=\int\left(\sqrt{\tan x}-\sqrt{\cot x}\right)\,\mathrm dx =\sqrt2\int\frac{\sin x-\cos x}{\sqrt{\sin2x}} \,\mathrm dx\]
\[ =-\sqrt2\int\frac{(\sin x+\cos x)'}{\sqrt{(\sin x+\cos x)^2-1}}\,\mathrm dx =-\sqrt2\int\frac{\mathrm du}{\sqrt{u^2-1}}\]
\[=-\sqrt2\cosh^{-1}(\sin x+\cos x) \]
Thus we get
\[A=\int\sqrt{\tan x} dx=\frac{(A-B)+(A+B)}2\]
\[= \frac{\sqrt2}2(\sin^{-1}(\sin x-\cos x)-\cosh^{-1}(\sin x+\cos x)) + C.\]
Note than you also get
\[B=\int\sqrt{\cot x} dx=-\frac{(A-B)-(A+B)}2\]
\[=- \frac{\sqrt2}2(\sin^{-1}(\sin x-\cos x)-\cosh^{-1}(\sin x+\cos x)) + C.\]
- answered
- 1545 views
- $3.00
Related Questions
- Integrate $\int \frac{1}{x^2+x+1}dx$
- Solve the inequality $x<\frac{6}{x-1}$
- Explain why does gradient vector points in the direction of the steepest increase?
-
MCQ Project 1:
6 Integration and Accumulation of Change
7 Differential Equations
8 Applications of Integration - Derivatives
- Find the limit as x-->0 for y = (e^x- 1)/[sin(nx)]
- Does $\lim_{n \rightarrow \infty} \frac{2^{n^2}}{n!}$ exist?
- Parametric, Polar, and Vector-Valued Equations for Kav10