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This is a trigonometric integration using trig identities.

S tanX^3 secX dX

S tanX^2 secX tanX dX

S (secX^2 -1) secX tanX dX

u = secX

du = secX tanX

S ( u^2 - 1) du

1/3secX^3 - secX + C

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โˆ™ 2010-11-22 02:47:47
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Q: What is the integral of tan cubed x secx dx?
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Related questions

What is the derivative of secxtanx?

d/dx(uv)=u*dv/dx+v*du/dxd/dx(secxtanx)=secx*[d/dx(tanx)]+tanx*[d/dx(secx)]-The derivative of tanx is:d/dx(tan u)=[sec(u)]2*d/dx(u)d/dx(tan x)=[sec(x)]2*d/dx(x)d/dx(tan x)=[sec(x)]2*(1)d/dx(tan x)=(sec(x))2=sec2(x)-The derivative of secx is:d/dx(sec u)=[sec(u)tan(u)]*d/dx(u)d/dx(sec x)=[sec(x)tan(x)]*d/dx(x)d/dx(sec x)=[sec(x)tan(x)]*(1)d/dx(sec x)=sec(x)tan(x)d/dx(secxtanx)=secx*[sec2(x)]+tanx*[sec(x)tan(x)]d/dx(secxtanx)=sec3(x)+sec(x)tan2(x)


What is the integral of tanx times sqrt secx dx?

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Integration of tan pow4x?

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Intergrate sec x?

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Integral of 1 divided by sinx cosx?

Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C


Integral of tan square x secant x?

convert tan^2x into sin^2x/cos^2x and secant x into 1/cos x combine terms for integral sin^2x/cos^3x dx then sub in u= cos^3x and du=-2sin^2x dx


What is the integral of tan x-3 dx?

In this specific example one would need to use the u substitution method. * Set u to be x - 3 * Derive x - 3 * u = x - 3 * du = dx Now that we have integrated u we can remove the x - 3 and substitute in u and remove the dx and substitute in du. This is what we have after substituting: * (the integrand of) tan(u)du Now integrate tan(u)du * the Integral of tan(u)du is: * sec2(u) Now resubstitute what we set as u. In this case we set x - 3 to u. This will give us our final answer and integral of tan(x-3)dx. * sec2(x - 3)


Integration of square root tan x dx?

for solving this ..the first thing to do is substitute tanx=t^2 then x=tan inverse t^2 then solve the integral..


What is the integral of tan squared x?

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How do you integrate sec cube 2x?

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What is the integral of 1 divided by the cosine squared of x with respect to x?

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What is the integral of sec squared x?

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What is the integration of tanx?

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How does one find the integral of secant cubed of x dx?

Find I = ∫ sec³ x dx. The answer is I = ½ [ log(sec x + tan x) + sec x tan x ]. * Here is how we may find it: Letting s = sec x, and t = tan x, we have, s² = 1 + t², dt = s² dx = (1 + t²) dx, and ds = st dx. Then, we obtain, dI = s³ dx = s dt. * Now, d(st) = s dt + t ds = dI + t ds = dI + st² dx = dI + s(s² - 1)dx = dI + s³ dx - s dx = 2dI - s dx; whence, 2dI = s dx + d(st). * Also, we have, s = (s² + st) / (s + t), whence s dx = (s² + st) dx / (s + t) = (dt + ds) / (s + t) = d(s + t) / (s + t) = d log(s + t). This gives us, 2dI = d log(s + t) + d(st). Integrating, we easily obtain, I = ½ [ log(s + t) + st ], which is the answer we sought. * Checking that we have arrived at the correct answer, we differentiate back: d(st) / dx = (st)'= st' + ts' = s³ + st² = 2s³ - s. d log(s + t) / dx = log'(s + t) = (s + t)' / (s + t) = (st + s²) / (s + t) = s. Thus, 2I' = [ st + log(s + t) ]' = 2s³; and I' = ½ [ st + log(s + t) ]' = s³, confirming that our answer is correct.