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If `y=x^(2)sin(x^(3))`, then `int ydx` will be :A. `-cos(x^(3))+C`B. `(-cos.(x^(3))/(3))+C`C. `cos(x^(3))+C`D. `((cos x^(3))/(3))+C` |
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Answer» Correct Answer - B `y=x_(2) sin x_(3)` `int x^(2) sin x^(3) dx` let `u=x_(3) rArr du = 3x_(2) dx` `int(sin xdx)/(3)=(1)/(3)(-cosx)+C=-(1)/(3) cosx_(3) +C`. |
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