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The anti-derivative of sqrt(x) :
sqrt(x)=x^(1/2)
The anti-derivative is x^(1/2+1) /(1/2+1) = (2/3) x^(3/2)
The anti-derivative is 4e^x is 4 e^x ( I hope you meant e to the power x)
The anti-derivative of -sin(x) is cos(x)
Adding, the anti-derivative is
(2/3) x^(3/2) + 4 e^x + cos(x) + C
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Which is the antiderivative of sinxcosx?
Using u-substitution (where u = sinx), you'll find the antiderivative to be 0.5*sin2x + C.
How do you solve 1 minus cosx divided by sinx plus sinx divided by 1 minus cosx to get 2cscx?
(1-cosx)/sinx + sinx/(1- cosx) = [(1 - cosx)*(1 - cosx) + sinx*sinx]/[sinx*(1-cosx)] = [1 - 2cosx + cos2x + sin2x]/[sinx*(1-cosx)] = [2 - 2cosx]/[sinx*(1-cosx)] = [2*(1-cosx)]/[sinx*(1-cosx)] = 2/sinx = 2cosecx
What is the antiderivative of 9sinx?
The antiderivative of 9sinx is simply just -9cosx. It is negetive because the derivative of cosx should have been -sinx, however, the derivative provided is positive. Therefore, it means that there should be a negative with cosx in order to make that sinx postive. (negative times negative eguals positive)
What is the derivative of sin x minus cos x?
d/dx(sinx-cosx)=cosx--sinx=cosx+sinx
What is the anti derivative of the square root of 1-x2?
-1
How do you prove the following equation the quantity of sin theta divided by 1 minus cos theta minus the quantity 1 plus cos theta divided by sin theta equals 0?
You will have to bear with the angle being represented by x because this browser will not allow characters from other alphabets!sin^2x + cos^2x = 1=> sin^2x = 1 - cos^x = (1 + cosx)(1 - cosx)Divide both sides by sinx (assuming that sinx is not zero).=> sinx = (1 + cosx)(1 - cosx)/sinxDivide both sides by (1 - cosx)=> sinx/(1 - cosx) = (1 + cosx)/sinx=> sinx/(1 - cosx) - (1 + cosx)/sinx = 0
Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?
2
What is the antiderivative 7x5-Cos x?
If the first term is 7x^5, ∫7x^5 -cox dx is the expression. You can split this up into two integrals if that helps you visualize the terms. ∫7x^5dx - ∫cox dx. We know that the antiderivative of cosx is sinx, so that is our second term. In the first term, we must undo the power rule, adding one to the power and multiplying by the reciprocal of the power. This gives us (7/6)x^6. So, our final antiderivative expression is (7/6)x^6-sinx+C, with C being an arbitrary constant.
Parenthesis 1 plus tanx end parenthesis divided by sinx equals cscx plus secx?
(1 + tanx)/sinxMultiply by sinx/sinxsinx + tanxsinxDivide by sin2x (1/sin2x) = cscxcscx + tan(x)csc(x)tanx = sinx/cosx and cscx = 1/sinxcscx + (sinx/cosx)(1/sinx)sinx cancels outcscx + 1/cosx1/cosx = secxcscx + secx
How do you integrate sinx divide sinx plus cosx?
1/2(x-ln(sin(x)+cos(x)))
What is sinx plus cosx plus cotx simplified?
There is no sensible or useful simplification.
What will be Limit x tends to 0 x square minus x plus 1 whole divide bi x minus sinx?
It is not possible to answer the question because it is ambiguous. The expression could be(x^2 - x + 1) / x - sinxor(x^2 - x + 1) / (x - sinx)Clearly, the answer will differ.This sort of ambiguity would have been avoided by using brackets.
