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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 formula of a raised to x?
integral (a^x) dx = (a^x) / ln(a)
What is the integral of -3dx?
∫(-3)dx = -3x + C
How do you integrate 2x over 2x-2?
Integral of 2x dx /(2x-2) Let 2x=u 2 dx = du dx = (1/2) du Integral of 2x dx /(2x-2) = Integral of (1/2) u du / (u-2) = Integral of 1/2 [ (u-2+2) / (u-2)] dx = Integral of 1/2 [ 1+ 2/(u-2)] dx = u/2 + (1/2) 2 ln(u-2) + C = u/2 + ln(u-2) + C = (2x/2) + ln(2x-2) + C = x + ln(2x-2) + C
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.
What is the integral of tanx times sqrt secx dx?
See related link below for answer
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)
Integration by parts of x tanx?
XtanX dx formula uv - int v du u = x du = dx dv = tanX dx v = ln(secX) x ln(secX) - int ln(secx) dx = X ln(secx) - x ln(secx) - x + C -----------------------------------------
Intergrate sec x?
Sec x dx = sec x (secx + tanx)/ (secx + tanx) dx . therefore the answer is ln |secx + tanx|
Integration of tan pow4x?
integral of (tanx)^4 (tanx)^4 = (tanx)^2 (tanx)^2 =(sec^2 x - 1)(tan^2 x) =(sec^2 x)(tan^2 x) - tan^2 x = integral of sec^2 x tan^2 x dx - integral of tan^2 x dx First, integral of sec^2 x tan^2 x dx Let u = tanx because that would make du = sec^2 x dx so then we have integral of u^2 du which is (1/3)u^3 substituting back in tanx we get (1/3)tan^3 x Next, integral of tan^2 x tan^2 x = sec^2 x -1 integral of sec^2 x - 1 = integral of sec^2 x dx - integral 1 dx = tanx - x so putting it all together we have integral of tan^4 x dx = (1/3)tan^3 x - tanx + x + C
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
What is the integral of tan squared x?
Note that for sec²(x) - tan²(x) = 1, we have: -tan²(x) = 1 - sec²(x) tan²(x) = sec²(x) - 1 Rewrite the expression as: ∫ (sec²(x) - 1) dx = ∫ sec²(x) dx - ∫ 1 dx Finally, integrate each expression to get: tan(x) - x + K where K is the arbitrary constant
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)
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
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..
How do you integrate sec cube 2x?
The integral of sec(x) is ln|secx+tanx| + C Since the derivative is taken to the third power, we have to consider the chain rule; the original equation must be to the fourth power, and in order for that to be canceled out, the equation must also have had a coefficient of 1/4. 2x is also subject to the chain rule. I would suggest u substitution. integral(sec(2x))^3 dx u=2x du=2dx dx=1/2du integral (sec(u))^3 *1/2 du 1/8 secxtanx + 1/8(ln|secx+tanx|^4) + C
What is the integral of the tangent of x with respect to x?
∫ tan(x) dx = -ln(cos(x)) + C C is the constant of integration.
