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OpenStudy (anonymous):
integral of x^3//(1-x^2) i got trig substitution to be me to the integral of sine thetea cubed. after that I would take one of the sines and convert. I would take sine to the second power. I changed this to 1-cos^2. Now I am stuck. I have the integral of (1-cos^2x)/(sin^x). any input?
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hartnn (hartnn):
is your original Q this : \(\int \dfrac{x^3}{\sqrt{1-x^2}}dx\)
hartnn (hartnn):
if you put x= sin t there, you'd get \(\int \sin^3 t \: dt\) and you can use the sin 3x formula. sin 3x = 3sin x -4 sin^3 x isolate sin^3 x from here and substitute in your integral.
OpenStudy (raden):
alternative : we can solve by u subs .. let u=1-x^2 ---> du = -2x dx or x dx = -du/2 x^2 = 1-u so, ur integral can be : |dw:1360314414033:dw|
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