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What is the anti-derivative of 4cosx divided by sinx to the power of 2?
Wiki User
∙ 14y ago
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Giles Wiegand
Wiki User
∙ 14y agou= sinx du=cosx 4 du/u^2 4u^-2du 4u^-1/-1 -4/u -4/sinx
answer is -4/sinx
the derivative of -4/sinx give you the original function 4cosx/(sinx)^2
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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
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 break 1 sinx divided 1-cosx?
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Is 1 plus sinX divided by 1 plus cscX equal to sinX?
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Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?
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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.
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.
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 anti derivative of the square root of 1-x2?
-1
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 derivative of 1 divided by sinx?
y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx
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 break 1 sinx divided 1-cosx?
0
Is 1 plus sinX divided by 1 plus cscX equal to sinX?
2
What is the derivative of sin x to the e to the xth power?
y = (sinx)^(e^x) ln(y) = ln((sinx)^(e^x)) ln(y) = (e^x)ln(sinx) (1/y)dy = (e^x)(1/sinx)(cosx)+ln(sinx)(e^x)dx (1/y)dy = (e^x)(cotx)+ln(sinx)(e^x)dx dy = ((sinx)^(e^x))((cotx)(e^x)+ln(sinx)(e^x))dx dy = ((e^x)(sinx)^(e^x))(cotx+ln(sinx))dx
Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?
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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
