Question

integrate 1/(R sq + × sq) power 3/2​

Answer 1

Answer:

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Answer 2

Answer:

hey mate here is your answer

e want to express (a2+u2)3/2(a2+u2)3/2 as something without square roots. We want to use some form of the Pythagorean trigonometric identity sin2x+cos2x=1sin2⁡x+cos2⁡x=1. Multiplying each side by a2cos2xa2cos2⁡x, we get a2tan2x+a2=a2sec2xa2tan2⁡x+a2=a2sec2⁡x, which is in the desired form. of (sum of two squares) = (something squared).

This suggests that we should use the substitution u2=a2tan2xu2=a2tan2⁡x. Equivalently, we substitute u=atanxu=atan⁡x and du=asec2xdxdu=asec2⁡xdx. Then

∫du(a2+u2)3/2=∫asec2xdx(a2+a2tan2x)3/2.

∫du(a2+u2)3/2=∫asec2⁡xdx(a2+a2tan2⁡x)3/2.

Applying the trigonometric identity considered above, this becomes

∫asec2xdx(a2sec2x)3/2=∫dxa2secx=1a2∫cosxdx,

∫asec2⁡xdx(a2sec2⁡x)3/2=∫dxa2sec⁡x=1a2∫cos⁡xdx,

which can be easily integrated as

=1a2sinx.

=1a2sin⁡x.

Since we set u=atanxu=atan⁡x, we substitute back x=tan−1(ua)x=tan−1⁡(ua) to get that the answer is

=1a2sintan−1ua.

=1a2sin⁡tan−1⁡ua.

Since sintan−1z=zz2+1√sin⁡tan−1⁡z=zz2+1, this yields the desired result of

Explanation:

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