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  • Math
    Hello! Probably we need to find all the terms of this progression. Recall that each next term of a geometric progression is obtained by multiplying the previous term by the quotient, denote it as...

    Asked by bleardd on via web

    1 educator answer.

  • Math
    Given `dy/dx = 1/sqrt(4-x^2) ` and `y(0) = pi` We have to find y. So we can write, `dy=dx/sqrt(4-x^2)` Integrating both sides we have, `y=int dx/sqrt{4-x^2)+C` Now let `x=2sint ` So, `dx=2cost...

    Asked by enotes on via web

    1 educator answer.

  • Math
    Complete the square at the denominator: `x^4 + 2x^2 + 2 = (x^2)^2 + 2x^2 + 1 + 1 = (x^2 + 1)^2 + 1.` Now we see the substitution `y = x^2 + 1,` then `dy = 2x dx,` and the integral becomes `int...

    Asked by enotes on via web

    1 educator answer.

  • Math
    We have to evaluate the integral: `\int \frac{1}{(x-1)\sqrt{x^2-2x}}dx` We can write the integral as: `\int \frac{1}{(x-1)\sqrt{x^2-2x}}dx=\int \frac{1}{(x-1)\sqrt{(x-1)^2-1}}dx` Let `x-1=t` So...

    Asked by enotes on via web

    1 educator answer.

  • Math
    By completing the square and making simple substitution, we will reduce this integral to a table one. `-x^2-4x = -(x^2 + 4x + 4) + 4 = -(x+2)^2 + 4 = 4 - (x+2)^2.` Now make a substitution `y =...

    Asked by enotes on via web

    1 educator answer.

  • Math
    To evaluate the given integral: `int_(-2)^(2)(dx)/(x^2+4x+13)` , we follow the first fundamental theorem of calculus: If f is continuous on closed interval [a,b], we follow: ` int_a^bf(x)dx =...

    Asked by enotes on via web

    1 educator answer.

  • Math
    Recall that `(arccos(x))' = -1/sqrt(1-x^2)` and make the substitution `y = arccos(x),` then `dy = -1/sqrt(1-x^2).` The limits of integration are from `arccos(0) = pi/2` to `arccos(1/sqrt(2)) =...

    Asked by enotes on via web

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  • Math
    We have to evaluate the definite integral: `\int_{0}^{1/\sqrt{2}}\frac{arc sinx}{\sqrt{1-x^2}}dx` Let `t= arcsinx` Differentiating both sides we get, `\frac{1}{\sqrt{1-x^2}}dx=dt`...

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    1 educator answer.

  • Math
    Make the substitution `u = sin(x),` then `du = cos(x) dx.` The integration limits for `u` are from `sin(0) = 0` to `sin(pi/2) = 1,` and the integral becomes `int_0^1 (du)/(1 + u^2) = arctan(1) -...

    Asked by enotes on via web

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  • Math
    We have to evaluate the integral: `\int_{\pi/2}^{\pi}\frac{sinx}{1+cos^2x}dx` Let `cosx=u` So, `-sinx dx=du` When `x=\pi/2, u=0` `x=\pi, u=-1` So we have,...

    Asked by enotes on via web

    1 educator answer.

  • Math
    Make the substitution `y = e^(-x),` then `dy = -e^(-x) dx` and `e^(-2x) = y^2.` The limits of integrations for `y` become from `e^(-ln2) = 1/e^(ln2) = 1/2` to `e^(-ln4) = 1/e^(ln4) = 1/4.` The...

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  • Math
    For the given integral problem:` int_0^(ln(5))e^x/(1+e^(2x))dx` , it resembles the basic integration formula for inverse tangent: `int_a^b (du)/(u^2+c^2) = (1/c)arctan(u/c) |_a^b` where we let:...

    Asked by enotes on via web

    1 educator answer.

  • Math
    To be able to solve for definite integral, we follow the first fundamental theorem of calculus: `int_a^b f(x) dx = F(x) +C` such that f is continuous and F is the antiderivative of f in a closed...

    Asked by enotes on via web

    1 educator answer.

  • Math
    Make the substitution `u = sqrt(4x^2 - 9),` then `du = (4x)/sqrt(4x^2 - 9) dx.` Inversely, `dx =sqrt(4x^2 - 9)/(4x) du = u/(4x) du` and `4x^2 = u^2 + 9.` The limits of integration become from...

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  • Math
    Make a substitution `y = 2x,` then `dy = 2 dx` and the limits of integration for `y` are from `0` to `3.` The integral becomes the table one: `int_0^3 1/(1+y^2) (dy)/2 = 1/2 (tan)^(-1)(y) |_(y=0)^3...

    Asked by enotes on via web

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  • Math
    We have to evaluate the integral : `\int_{0}^{\sqrt{2}}\frac{dx}{\sqrt{4-x^2}}` let `x=2sin t` So, `dx=2cos t dt` When x=0, t=0 and when `x=\sqrt{2}, t=\pi/4` So we have,...

    Asked by enotes on via web

    1 educator answer.

  • Math
    Make a substitution `x = 3y,` then `dx = 3 dy` and the limits of integration for y are from 0 to 1/18. The integral becomes a table one: `int_0^(1/18) (9 dy)/(sqrt(9-9y^2)) = int_0^(1/18) (3...

    Asked by enotes on via web

    1 educator answer.

  • Math
    We have to evaluate the integral:`\int \frac{x-2}{(x+1)^2+4}dx` Let `x+1=u` So, `dx=du` Hence we have, `\int \frac{x-2}{(x+1)^2+4}dx=\int \frac{u-3}{u^2+4}du` `=\int...

    Asked by enotes on via web

    1 educator answer.

  • Math
    This integral may be taken by dividing into two parts: `int (x dx)/(x^2 + 1) - 3 int (dx)/(x^2 + 1).` The second integral is `3arctan(x) + C.` The first integral requires the substitution `u =...

    Asked by enotes on via web

    1 educator answer.

  • Math
    We have to evaluate the integral: `\int\frac{sinx}{7+cos^2x}dx` Let `cos x=u` So, `-sinxdx=du` Hence we have, `\int \frac{sinx}{7+cos^2x}dx=\int\frac{-du}{u^2+7}` `=-\int...

    Asked by enotes on via web

    1 educator answer.

  • Math
    We have to evaluate the integral : `\int \frac{sec^2x}{\sqrt{25-tan^2x}}dx` Let `tanx =t` So, `sec^2x dx=dt` Therefore we have, `\int \frac{sec^2x}{\sqrt{25-tan^2x}}dx=\int...

    Asked by enotes on via web

    1 educator answer.

  • Math
    Given the integral: `\int \frac{2}{x\sqrt{9x^2-25}}dx` Let `x=\frac{5}{3}sect` `` So, `dx=\frac{5}{3}sect tant dt` Hence we have, `\int \frac{2}{x\sqrt{9x^2-25}}dx=\int \frac{\frac{10}{3}sect...

    Asked by enotes on via web

    1 educator answer.

  • Math
    We have to evaluate the integral: `\int \frac{1}{x\sqrt{1-(ln x)^2}}dx` Let `ln x=u` So, `\frac{dx}{x}=du` Therefore we have, `\int \frac{1}{x\sqrt{1-(ln(x))^2}}dx=\int \frac{du}{\sqrt{1-u^2}}`...

    Asked by enotes on via web

    1 educator answer.

  • Math
    Make the substitution `u = t^2/5,` then `du = (2t dt)/5,` `t dt = 5/2 du` and `t^4 = 25u^2.` The integral becomes `int (5/2)/(25u^2 + 25) du = 1/10 int (du)/(1 + u^2) = 1/10 arctan(u) + C =...

    Asked by enotes on via web

    1 educator answer.

  • Math
    Indefinite integral are written in the form of int `f(x) dx = F(x) +C` where: f(x) as the integrand F(x) as the anti-derivative function C as the arbitrary constant known...

    Asked by enotes on via web

    1 educator answer.

  • Math
    Given the integral : `\int \frac{1}{4+(x-3)^2}dx` let `x-3=t` So, `dx=dt` therefore we have, `\int \frac{1}{4+(x-3)^2}dx=\int \frac{1}{4+t^2}dt` `=\int \frac{1}{2^2+t^2}dt`...

    Asked by enotes on via web

    1 educator answer.

  • Math
    Indefinite integral are written in the form of `int f(x) dx = F(x) +C` where: f(x) as the integrand F(x) as the anti-derivative function C as the arbitrary constant known...

    Asked by enotes on via web

    1 educator answer.

  • Math
    Recall that the indefinite integral is denoted as: `int f(x) dx =F(x)+C` There properties and basic formulas of integration we can apply to simply certain function. For the problem `int...

    Asked by enotes on via web

    1 educator answer.

  • Math
    We have to evaluate `\int \frac{dx}{\sqrt{1-4x^2}}` Let `x=\frac{1}{2} sint ` So, `dx= \frac{1}{2}cost dt` Hence we have, `\int \frac{dx}{\sqrt{1-4x^2}}=\int \frac{\frac{1}{2}cost...

    Asked by enotes on via web

    1 educator answer.

  • Math
    We have to evaluate the integral: `\int \frac{dx}{\sqrt{9-x^2}}` let `x=3sint` So, `dx=3cost dt` Hence we have, `\int \frac{dx}{\sqrt{9-x^2}}=\int \frac{3cost}{\sqrt{9-9sin^2t}}dt`...

    Asked by enotes on via web

    1 educator answer.

  • Math
    First, check that the given point satisfies the equation: `arctan(1 + 0) = 0 + pi/4` is true. The slope of the tangent line is `y'(x)` at the given point. Differentiate the equation with respect to...

    Asked by enotes on via web

    1 educator answer.

  • Math
    This function is defined on entire real axis and is differentiable everywhere. Its derivative is `f'(x) = 1/(1+x^2) - 1/(1 + (x-4)^2) = (1 + (x-4)^2 - 1 - x^2)/((1+x^2)(1 + (x-4)^2)) =` `= (-8x +...

    Asked by enotes on via web

    1 educator answer.

  • Math
    The derivative of y with respect to is denoted as` y'` or `(dy)/(dx)` . For the given equation: `y = arctan(x/2) -1/(2(x^2+4))` , we may apply the basic property of derivative: `d/(dx) (u-v)...

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    1 educator answer.

  • Math
    Recall that the derivative of y with respect to is denoted as `y'` or `(dy)/(dx)` . For the given equation: `y = arctan(x) +x/(1+x^2)` , we may apply the basic property of derivative: `d/(dx)...

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    1 educator answer.

  • Math
    We need the derivatives of `ln(u)` and `arctan(u),` they are `1/u` and `1/(1 + u^2),` and the chain rule. The result is `y'(t) = (2t)/(t^2+4) - 1/2*(1/2)/(1+t^2/4) =(2t)/(t^2+4) - 1/(t^2 + 4) = (2t...

    Asked by enotes on via web

    1 educator answer.

  • Math
    `y = 2arccos(x) - 2sqrt(1-x^2)` First, express the radical in exponent form. `y = 2arccos(x)-2(1-x^2)^(1/2)` To take the derivative of this, use the following formulas: `d/dx(arccos(u)) =...

    Asked by enotes on via web

    1 educator answer.

  • Math
    Recall that the derivative of a function f at a point x is denoted as f'(x). The given function: `f(x)= arcsin(x)+arccos(x)` has inverse trigonometric terms. There are basic formulas for the...

    Asked by enotes on via web

    1 educator answer.

  • Math
    This function is a composite one, it may be expressed as `h(t) = u(v(t)),` where `v(t) = arccos(t)` and `u(y) = sin(y).` The chain rule is applicable here, `h'(t) = u'(v(t))*v'(t).` This gives us...

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    1 educator answer.

  • Math
    We use the product rule, `(uv)' = u'v + uv',` for `u = x^2` and `v = arctan(5x),` and then the chain rule: `h'(x) = 2x*arctan(5x) + x^2 (arctan(5x))' =` `= 2x*arctan(5x) + (5 x^2)/(1 + 25x^2).`

    Asked by enotes on via web

    1 educator answer.

  • Math
    The given function `g(x)` has a form `u(x)/v(x),` thus its derivative may be determined by the quotient rule: `(u/v)' = (u' v - u v')/v^2.` Also we need to know that the derivative of acrsine...

    Asked by enotes on via web

    1 educator answer.

  • Math
    The given function `f(x) = arctan(sqrt(x))` is in a inverse trigonometric form. The basic derivative formula for inverse tangent is: `d/(dx) arctan(u) = ((du)/(dx))/sqrt(1-u^2)` . Using...

    Asked by enotes on via web

    1 educator answer.

  • Math
    The given function: `f(x) =arctan(e^x)` is in a form of inverse trigonometric function. It can be evaluated using the derivative formula for inverse of tangent function: `d/(dx)arctan(u) =...

    Asked by enotes on via web

    1 educator answer.

  • Math
    The given function: `f(x) =arcsec(2x) ` is in a form of an inverse trigonometric function. For the derivative formula of an inverse secant function, we follow:...

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    1 educator answer.

  • Math
    To take the derivative of the given function:` g(x) =3arccos(x/2)` , we can apply the basic property: `d/(dx) [c*f(x)] = c * d/(dx) [f(x)]` . then `g'(x) = 3 d/(dx) (arccos(x/2))` To solve for...

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  • Math
    The derivative of a function with respect to t is denoted as f'(t) The given function:` f(x) = arcsin(t^2) ` is in a form of a inverse trigonometric function. Using table of derivatives, we have...

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    1 educator answer.

  • Math
    The derivative of function f with respect to x is denoted as` f'(x)` . To take the derivative of the given function: `f(x) =2arcsin(x-1)` , we can apply the basic property: `d/(dx) [c*f(x)] = c *...

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    1 educator answer.

  • Math
    `arcsin(-x)=-arcsin(x)` Let `y=arcsin(-x)` `sin(y)=sin(arcsin(-x))` `sin(y)=-x` `-sin(y)=x` `sin(-y)=x` `arcsin(sin(-y))=arcsin(x)` `-y=arcsin(x)` `y=-arcsin(x)` Therefore:...

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  • Math
    The function at the left side, `arccos(x),` is defined at `[-1, 1],` because it is the range of `cos(x).` The function at the right side, `arcsec(x),` is defined at `(-oo, -1] uu [1, +oo],`...

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    1 educator answer.

  • Math
    `arcsin (sqrt(2x)) = arccos(sqrtx)` To solve, let's consider the right side of the equation first. Let it be equal to `theta`. `theta = arccos(sqrtx)` Then, express it in terms of cosine. `cos...

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  • Math
    The range of arctan function is `(-pi/2, pi/2),` on which interval tangent function is strictly monotone. Therefore we may take tangent on both sides, and the solution set will remain the same....

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