# Homework Help

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

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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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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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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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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 +...