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  • 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 =...

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  • 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...

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

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  • 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
    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) -...

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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 arc sinx=t Differentiating both sides we get, `\frac{1}{\sqrt{1-x^2}}dx=dt`...

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  • 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)) =...

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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 `x=\sqrt{2}, t=\pi/4` So we have,...

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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...

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  • 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
    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...

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  • 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 obviously `3arctan(x) + C.` The second requires the substitution `u =...

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  • 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...

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  • 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...

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  • 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...

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  • 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`...

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  • 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 =...

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  • 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...

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  • 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...

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  • 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`...

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  • 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...

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  • 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...

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  • 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 +...

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  • 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...

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  • 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).`

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  • 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. We can solve for the derivative of...

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  • 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...

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  • 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 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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  • 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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  • 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 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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  • 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) =...

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  • Math
    Given: `arctan(sqrt3/3)` `tany=sqrt3/3` `y=pi/6` when y is in the the domain `-pi/2<y<pi/2` Final Answer: `y=pi/6,{-pi/2<y<pi/2}`

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  • Math
    For a negative number `x` (actually, for `x` less than or equal to `-1` ) its arcsecant is such number `y` in `(pi/2, pi]` that `sec(y) = x.` This number is unique. This is the same as `cos(y) =...

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  • Math
    Given `sin(arctan(3/4))` Imagine a right triangle with the opposite leg with the length of 3 and an adjacent leg with the length of 4. The length of the hypotenuse can be found using the...

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  • Math
    `tan(arccos(sqrt2/2))_` `cosy=sqrt2/2` Imagine a right triangle with the adjacent side of length `sqrt2` and the hypotenuse of length 2. Using the the pythagorean theorem `(sqrt2)^2+x^2=2^2`...

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  • Math
    `cot(arcsin (-1/2))` To evaluate this, let's first consider the innermost expression. Let it be equal to y. `y = arcsin(-1/2)` Rewriting it in terms of sine function, the equation becomes `sin (y)...

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  • Math
    Given: `sec(arctan(-3/5))` `tany=-3/5` when y is in the domain `-pi/2<y<0` Imagine a right triangle with opposite side having a length of 3 and adjacent side having a length of 5. The...

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  • Math
    The range of arcsine function is `[-pi/2, pi/2],` 1/2 belongs to this interval and sine is monotone on this interval. Therefore we may apply sine for both parts and don't lose any solutions or get...

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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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  • Math
    `arcsin(1/2)` Let this expression be equal to y. `y =arcsin(1/2)` Rewriting it in terms of sine function, the equation becomes: `sin(y) = 1/2` Base on the Unit Circle Chart (see attached figure),...

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  • Math
    `arcsin(0)` Let this expression be equal to y. `y =arcsin(0)` Re-writing this equation in terms of sine function, it becomes: `sin (y) = 0` Base on the Unit Circle Chart (see attached figure), sine...

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  • Math
    `arccos(1/2)` Let this expression be equal to y. `y = arccos(1/2)` Rewriting this in terms of cosine function the equation becomes: `cos(y) =1/2` Base on the Unit Circle Chart, cosine is 1/2 at...

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  • Math
    `arccos(1)` Let this expression be equal to y. `y = arccos(1)` Rewriting it in terms of cosine function, the equation becomes: `cos(y) = 1` Base on the Unit Circle Chart, cosine is 1 at angles 0...

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  • Math
    `int(x^4+5^x)dx` `=x^5/5+5^x/ln5+C`

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  • Math
    Recall the First Fundamental Theorem of Calculus: If f is continuous on closed interval [a,b], we follow: `int_a^bf(x)dx` = F(b) - F(a) where F is the anti-derivative of f on [a,b]. This...

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  • Math
    To evaluate the integral: `int_(-4)^(4) 3^(x/4) dx` , we follow the formula based from the First Fundamental Theorem of Calculus: `int_a^bf(x)dx=F(b)- F(a)` wherein f is a continuous and F is...

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  • Math
    `y = log_10 (2x)` The line is tangent to the graph of function at (5,1). The equation of the tangent line is _______. To solve this, we have to determine the slope of the tangent. Take note that...

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  • Math
    To use logarithmic differentiation, take natural logarithm of both sides: `ln(y) = (x-1)ln(x).` Then differentiate this equation with respect to `x` and obtain `(y')/y = (x-1)/x + ln(x) = 1 - 1/x +...

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