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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)
What else can I help you with?
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..
What is the integration of tanx?
The integral of tan(x) dx = ln | sec(x) | + cto solve... tan(x) = sin(x)/cos(x)the integral of (sin(x)/cos(x) dx) ... let u = cos(x) then du = -sin(x) dx= the integral of (1/u -du)= -ln | u | + c= -ln | cos(x) | + c= ln | (cos(x))^-1 | + c ... or ... ln | 1/cos(x) | + c= ln | sec(x) | + c
Integral of -x 2?
The integral of -x2 is -1/3 x3 .
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 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
What is the integral of x3 ex4 dx?
∫x3ex4 dx = 1/4ex4 + c To solve, let y = x4, then: dy = 4x3 dx ⇒ 1/4dy = x3 dx ⇒ ∫x3ex4 dx =∫ex4 x3 dx = ∫ey 1/4 dy = 1/4ey + c but y = x4, thus: = 1/4ex4 + 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 indefinite integral?
An indefinite integral is a version of an integral that, unlike a definite integral, returns an expression instead of a number. The general form of a definite integral is: ∫ba f(x) dx. The general form of an indefinite integral is: ∫ f(x) dx. An example of a definite integral is: ∫20 x2 dx. An example of an indefinite integral is: ∫ x2 dx In the definite case, the answer is 23/3 - 03/3 = 8/3. In the indefinite case, the answer is x3/3 + C, where C is an arbitrary constant.
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
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..
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.
Integration of xtanx?
To integrate ( x \tan(x) ), we can use integration by parts. Let ( u = x ) and ( dv = \tan(x) , dx ). This gives ( du = dx ) and ( v = -\ln|\cos(x)| ). Applying the integration by parts formula ( \int u , dv = uv - \int v , du ), we obtain: [ \int x \tan(x) , dx = -x \ln|\cos(x)| + \int \ln|\cos(x)| , dx + C ] The integral ( \int \ln|\cos(x)| , dx ) does not have a simple closed form, so the final result may be expressed in terms of this integral along with the logarithmic term.
What is the integration of tanx?
The integral of tan(x) dx = ln | sec(x) | + cto solve... tan(x) = sin(x)/cos(x)the integral of (sin(x)/cos(x) dx) ... let u = cos(x) then du = -sin(x) dx= the integral of (1/u -du)= -ln | u | + c= -ln | cos(x) | + c= ln | (cos(x))^-1 | + c ... or ... ln | 1/cos(x) | + c= ln | sec(x) | + c
Integral of -x 2?
The integral of -x2 is -1/3 x3 .
What is the integral of 1 divided by the cosine squared of x with respect to x?
∫ 1/cos2(x) dx = tan(x) + C C is the constant of integration.
