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Evaluate the indefinite integral: xsin(x^2)dx We are supposed to use U substitution and i substituted everything in but I'm not sure I substituted du correctly.
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i would try using u = x^2 .
i did and then i made du=xsinsdx but when I put it all together it is not making sense
-0.5cos(x^2)
let u = x^2. du = 2x dx or x dx = du /2. then the integration will be sin (u) du/2
thank you both!
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welcome ;)
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