Guide :

how to findout integral lnx sin3x dx

by integration by parts

Research, Knowledge and Information :


What is the antiderivative of sin3x? | eNotes


Get an answer for 'What is the antiderivative of sin3x?' and find homework help for other ... we'll have to determine the indefinite integral of sin 3x. Int sin 3x dx.
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Integration by Parts the Integral of cos(ln(x)) - YouTube


May 28, 2015 · Integration by Parts the Integral of cos(ln(x)) Skip navigation ... Find out why Close. ... Integrate cos(ln x) dx - Duration: ...

Integration by parts of (xSin3x)dx - YouTube


Dec 30, 2011 · Integration by parts of (xSin3x)dx. Skip navigation Sign in. ... Find out why Close. ... Integral of ln(x)/x^2, ...

How do you find the integral of cos(3x) dx? | Socratic


How do you find the integral of #cos(3x) dx#? ... Find out the integration of csc^5x dx??? ... (3 ln(x)) #? Pre ...
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How do you find the integral of (sinx)^3 dx? | Socratic


How do you find the integral of #(sinx)^3 dx#? ... (sinx)^3=3sinx-sin3x=> (sinx)^3=3/4*sinx-1/4sin3x # ... Find out the integration of csc^5x dx???
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Integration by Parts - analyzemath.com


... Use integration by parts to evaluate the integral ln(x) dx Solution ... Use the table of integrals and the method of integration by parts to find the integrals ...
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calculus - How to Evaluate $\int e^{2x}\sin(3x)\ dx ...


... (3x)\ dx = \int e^{2x}\Im(\cos(3x)+i\sin(3x))\ dx =\int e^{2x}\Im(e ... (so it won't cancel out). Call the integral $I$ and integrate by parts twice to get ...
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Does the integral $\\int_{2}^{+ \\infty} \\frac{\\arctan(e^x ...


... \frac{\arctan(e^x)\sin3x}{\sqrt{\ln(x)}}dx$$ We have that: $$0 \leq \frac{\arctan(e^x)\sin3x ... How can I show if the integral converges or not? Thank you ...
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how do you find the integral of (cos^3x)/(sin^2x)dx?


how do you find the integral of (cos^3x)/(sin^2x) ... how do you find the integral of (cos^3x)/(sin^2x)dx? ... Working out the integral gives you ∫ cos ...
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Suggested Questions And Answer :


how to findout integral lnx sin3x dx


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120(ln(5√x^3)+ln(3√(y^2)-ln(4√(16z^5)

What do you want done with it? Here's some simplifying, but without an actual question we don't know what kind of answer we're looing for. 120( ln(5x^(3/2)) + ln(3y) - ln(4 * (16z^5)^(1/2) ) ) 120( ln(5) + ln(x^(3/2)) + ln(3) + lny - ln(4 * 4 * z^(5/2) ) ) 120( ln(5) + (3/2) lnx + ln(3) + lny - ln(16z^(5/2)) ) 120( ln(5) + (3/2) lnx + ln(3) + lny - ln(16) - ln(z^5/2) ) 120( ln(5) + (3/2) lnx + ln(3) + lny - ln(16) - (5/2) lnz ) 120 ln(5) + 180 lnx + 120 ln(3) + 120 lny - 120 ln(16) - 300 lnz 180 lnx + 120 lny - 300 lnz + 120 ln(5) + 120 ln(3) - 120 ln(16) 180 lnx + 120 lny - 300 lnz + 120(ln(5) + ln(3) - ln(16)) 180 lnx + 120 lny - 300 lnz + 120 ln(5*3/16) 180 lnx + 120 lny - 300 lnz + 120 ln(15/16)
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x^y=e^x-y , prove that dy/dx=lnx/(1+lnx)^2

Applying the operator ln^,
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Write an equation of the tangent line to the curve given by y=(lnx)^lnx, at x=e.

let u = ln(x).  Then for y = ln(x)^ln(x) we have y=u^u and the chain rule gives us dy/dx=dy/du*du/dx Assume u>0 (there are problems otherwise) then ln(y) = ln(u^u) =ln(u)*u Differentiate both sides using chain rule on the left and product rule on the right dy/du*1/y = ln(u)+1/u*u=ln(u)+1 or dy/du = (ln(u)+1)*y = (ln(u) + 1)*u^u = (ln(ln(x))+1)*ln(x)^ln(x) du/dx = d/du(ln(x)) = 1/x so the derivative is ((ln(ln(x))+1)*ln(x)^ln(x))/x The slope of the tangent is (ln(ln(e))+1)*ln(e)^(ln(e))/e = (ln(1)+1)(1^1)/e=1/e now y = 1 at x=e so y=mx+b is y=1/e(x)+1
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(sin x + sin 2x + sin 3x - sin 4x)/ (cos x + cos 2x + cos 3x + cos 4x) =

Should all the signs be positive?  I'll assume initially there are all pluses, so we have: (sinx+sin2x+sin3x+sin4x)/(cosx+cosx+cos3x+cos4x). The shortcut is to use the identity: sinA+sinB=2sin((A+B)/2)cos((A-B)/2) and cosA+cosB=2cos((A+B)/2)cos((A-B)/2). sinx+sin4x=2sin(5x/2)cos(3x/2); sin2x+sin3x=2sin(5x/2)cos(x/2); cosx+cos4x=2cos(5x/2)cos(3x/2); cos2x+cos3x=2cos(5x/2)cos(x/2). (sinx+...+sin4x)/(cosx+...+cos4x)=2sin(5x/2)(cos(3x/2)+cos(x/2))/2cos(5x/2)(cos(3x/2)+cos(x/2))=tan(5x/2). [This vast simplification is unlikely to apply if we have -sin4x instead of +sin4x, but we'll see where it leads: sinA-sinB=2cos((A+B)/2)sin((A-B)/2), so sin3x-sin4x=-2cos(7x/2)sin(x/2) and sinx-sin4x=-2cos(5x/2)sin(3x/2). So (sinx+sin2x+sin3x-sin4x)=-2cos(5x/2)sin(3x/2)+2sin(5x/2)cos(x/2)=2sin(3x/2)cos(x/2)-cos(7x/2)sin(x/2). more...]
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x^y=e^x-y prove dy/dx=lnx/(1+lnx)^2

Applying the operator ln^,
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prove: (sin5x+sin3x)/(sin5x-sin3x)=tan4x/tanx


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sin3x + sinx/cosx + cos3x= tan2x

sinA+sinB=2sin((A+B)/2)cos((A-B)/2); cosA+cosB=2cos((A+B)/2)cos((A-B)/2). If A=3x and B=x, sin3x+sinx=2sin2xcosx; cos3x+cosx=2cos2xcosx. Therefore (sin3x+sinx)/(cosx+cos3x)=2sin2xcosx/2cos2xcosx=tan2x QED.
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sin3x+sin5x-sinx-sin7x=sin3x


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what is the derivative of y= ([x^2/lnx])^3

[ 3(x^2/lnx)^2 ] [ (lnx)(2x)-((x^2)(1/x) ]                                    (lnx)^2
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