Evaluate the integral with hyperbolic or trigonometric substitution. ?

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Answer 1 · 2018-03-29T03:21:43

$\int \frac{1}{\sqrt{x^{6} - x^{4}}} d x = cos ({sec}^{- 1} x) + c$ $int1/sqrt(x^6-x^4)dx=cos(sec^-1x)+c$

$I = \int \frac{1}{\sqrt{x^{6} - x^{4}}} d x$ $I=int1/sqrt(x^6-x^4)dx$

Let

$x = sec t$ $x=sect$

$\frac{d x}{d t} = sec t tan t$ $dx/dt=sect tant$

$d x = sec t tan t d t$ $dx=sect tant dt$

$\sqrt{x^{6} - x^{4}} = \sqrt{{sec}^{6} t - {sec}^{4} t}$ $sqrt(x^6-x^4)=sqrt(sec^6t-sec^4t)$

$= \sqrt{{sec}^{4} t ({sec}^{2} t - 1)}$ $=sqrt(sec^4t(sec^2t-1))$

$\sqrt{x^{6} - x^{4}} = {sec}^{2} t tan t$ $sqrt(x^6-x^4)=sec^2t tant$

$I = \int \frac{1}{\sqrt{x^{6} - x^{4}}} d x$ $I=int1/sqrt(x^6-x^4)dx$

$I = \int \frac{sec t tan t d t}{{sec}^{2} t tan t}$ $I=int(sect tant dt)/(sec^2t tant)$

$I = \int cos t d t$ $I=intcostdt$

$I = sin t$ $I=sint$

$x = sec t$ $x=sect$

$t = {sec}^{- 1} x$ $t=sec^-1x$

$I = cos ({sec}^{- 1} x) + c$ $I=cos(sec^-1x)+c$

$\int \frac{1}{\sqrt{x^{6} - x^{4}}} d x = cos ({sec}^{- 1} x) + c$ $int1/sqrt(x^6-x^4)dx=cos(sec^-1x)+c$