Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 13362 by tawa tawa last updated on 19/May/17

∫ (x^3 /(√(16 − x^2 ))) dx

∫x316−x2dx

Answered by ajfour last updated on 19/May/17

I=(1/2)∫ ((x^2 (2xdx))/(√(16−x^2 ))) let x^2 =t ⇒ 2xdx=dt I=(1/2)∫((tdt)/(√(16−t))) =−(1/2)∫((16−t−16)/(√(16−t))) dt =−(1/2)∫(√(16−t)) dt−8∫((−dt)/(√(16−t))) =(1/2)[(2/3)(16−t)^(3/2) ]−8×2(√(16−t))+C as t=x^2 we have I=(1/3)(16−x^2 )^(3/2) −16(16−x^2 )^(1/2) +C .

I=12∫x2(2xdx)16−x2letx2=t⇒2xdx=dtI=12∫tdt16−t=−12∫16−t−1616−tdt=−12∫16−tdt−8∫−dt16−t=12[23(16−t)3/2]−8×216−t+Cast=x2wehaveI=13(16−x2)3/2−16(16−x2)1/2+C.

Commented by tawa tawa last updated on 19/May/17

God bless you sir.

Godblessyousir.

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/May/17

x=4sint⇒dx=4costdt cost=(√(1−(x^2 /(16))))=((√(16−x^2 ))/4) I=∫((4^3 sin^3 t.4costdt)/(4cost))=64∫sin^3 tdt =64∫sint(1−cos^2 t)dt= =64∫[sint−sintcos^2 t]dt= =64(−cost+(1/3)cos^3 t)+C= =64(−((√(16−x^2 ))/4)+(1/3).(((16−x^2 )(√(16−x^2 )))/4^3 ))+C =−(1/3)(32+x^2 )(√(16−x^2 ))+C .■

x=4sint⇒dx=4costdtcost=1−x216=16−x24I=∫43sin3t.4costdt4cost=64∫sin3tdt=64∫sint(1−cos2t)dt==64∫[sint−sintcos2t]dt==64(−cost+13cos3t)+C==64(−16−x24+13.(16−x2)16−x243)+C=−13(32+x2)16−x2+C.◼

Commented by ajfour last updated on 19/May/17

error in line #5

You can't use 'macro parameter character #' in math mode

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 19/May/17

thank you my friend.error fixed.

thankyoumyfriend.errorfixed.

Terms of Service

Privacy Policy

Contact: info@tinkutara.com