0-1-2-x-sin-1-x-2-1-x-4-dx-2-x-tanh-2-1-x-dx-

Question Number 115781 by bemath last updated on 28/Sep/20

\underset{0}{\int^{\frac{1}{\sqrt{2}}}} \frac{x \sin^{- 1} (x^{2})}{\sqrt{1 - x^{4}}} d x = ?

\int 2^{- x} \tanh (2^{1 - x}) d x = ?

Answered by bobhans last updated on 28/Sep/20

I = \underset{0}{\int^{\frac{1}{\sqrt{2}}}} \frac{x . \sin^{- 1} (x^{2})}{\sqrt{1 - x^{4}}} d x

l e t u = \sin^{- 1} (x^{2}) \to {\begin{cases} x = \frac{1}{\sqrt{2}} \to u = \frac{π}{6} \\ x = 0 \to u = 0 \end{cases}

I = \underset{0}{\int^{\frac{π}{6}}} \frac{1}{2} u d u = {[\frac{1}{4} u^{2}]}_{0}^{\frac{π}{6}}

= \frac{1}{4} \times \frac{π^{2}}{36} = \frac{π^{2}}{144}

Answered by Dwaipayan Shikari last updated on 28/Sep/20

\int_{0}^{\frac{1}{\sqrt{2}}} \frac{{xsin}^{- 1} (x^{2})}{\sqrt{1 - x^{4}}} dx

\frac{1}{2} \int_{0}^{\frac{1}{2}} \frac{\sin^{- 1} (t)}{\sqrt{1 - t^{2}}} dt

\frac{1}{4} {[{(\sin^{- 1} (t))}^{2}]}_{0}^{\frac{1}{2}} = \frac{1}{4} . {(\frac{π}{6})}^{2} = \frac{π^{2}}{144}

Answered by mathmax by abdo last updated on 28/Sep/20

changement \arcsin (x^{2}) = t give x^{2} = sint \Rightarrow x = \sqrt{sint} \Rightarrow

\int_{0}^{\frac{1}{\sqrt{2}}} \frac{xarcsin (x^{2})}{\sqrt{1 - x^{4}}} dx = \int_{0}^{\frac{π}{6}} \frac{t \sqrt{sint}}{\sqrt{1 - \sin^{2} t}} \times \frac{cost}{2 \sqrt{sint}} dt

= \frac{1}{2} \int_{0}^{\frac{π}{6}} tdt = \frac{1}{2} {[\frac{t^{2}}{2}]}_{0}^{\frac{π}{6}} = \frac{1}{4} \times \frac{π^{2}}{36} = \frac{π^{2}}{144}

Answered by Ar Brandon last updated on 28/Sep/20

I = \int 2^{- x} \tanh (2^{1 - x}) d x = - \frac{1}{2 \ln 2} \int - 2^{1 - x} \ln 2 \tanh (2^{1 - x}) d x

= - \frac{1}{2 \ln 2} \int \tanh (u) du {\frac{d}{d x} (2^{1 - x}) = - 2^{1 - x} \ln 2}

= - \frac{1}{2 \ln 2} {utanh (u) - \int \frac{u}{1 - u^{2}} du}

= - \frac{1}{2 \ln 2} {utanh (u) + \frac{\ln ∣ 1 - u^{2} ∣}{2}} + C

= - \frac{1}{2 \ln 2} {2^{1 - x} \tanh (2^{1 - x}) + \frac{\ln ∣ 1 - 2^{2 (1 - x)} ∣}{2}} + C