t-5-2-t-2-dt-

Question Number 120040 by bramlexs22 last updated on 28/Oct/20

\int \frac{t^{5}}{\sqrt{2 + t^{2}}} d t

Answered by john santu last updated on 28/Oct/20

l e t 2 + t^{2} = h^{2} \to t d t = h d h

\int \frac{t^{4} t d t}{\sqrt{2 + t^{2}}} = \int \frac{{(h^{2} - 2)}^{2} h d h}{h}

\int (h^{4} - 4 h^{2} + 4) d h = \frac{1}{5} h^{5} - \frac{4}{3} h^{3} + 4 h + c

= \frac{1}{15} \sqrt{2 + t^{2}} {3 {(2 + t^{2})}^{2} - 20 (2 + t^{2}) + 60} + c

= \frac{\sqrt{2 + t^{2}}}{15} {3 t^{4} - 8 t^{2} + 32} + c

Commented by Lordose last updated on 28/Oct/20

FASTEST

Commented by bramlexs22 last updated on 29/Oct/20

w a w \dots

Answered by Lordose last updated on 28/Oct/20

set t = \sqrt{2} tanx \Rightarrow dt = \sqrt{2} (1 + t^{2}) dx

\int \frac{{(\sqrt{2})}^{6} \tan^{5} {xsec}^{2} x}{\sqrt{2} secx} dx

4 \sqrt{2} \int \tan^{5} xsecxdx

4 \sqrt{2} \int tanxsecx {(\sec^{2} x - 1)}^{2} dx

u = secx \Rightarrow du = secxtanxdx

4 \sqrt{2} \int {(u^{2} - 1)}^{2} du

4 \sqrt{2} \int (u^{4} - {2 u}^{2} + 1) du

4 \sqrt{2} (\frac{u^{5}}{5} - \frac{{2 u}^{3}}{3} + u) + C

4 \sqrt{2} (\frac{\sec^{5} x}{5} - \frac{{2 \sec}^{3} (x)}{3} + secx) + C

x = \tan^{- 1} (\frac{t}{\sqrt{2}})

I = \frac{1}{15} \sqrt{t^{2} + 2} ({3 t}^{2} - 8 t + 32) + C

Commented by bramlexs22 last updated on 29/Oct/20

y e s