integral - the result of a mathematical integration; F(x) is the integral of f(x) if dF/dx = f(x) figuring, reckoning, calculation, computation - problem solving that involves numbers or quantities indefinite integral - the set of functions F(x) + C, where C is any real number, such...
Log In Sign Up Subjects Math Calculus Antiderivative Integration - e^sinx(x cosx - tanx.secx)dx Question:Integration - e^sinx(x cosx - tanx.secx)dx Indefinite Integration.The reversing process of differentiation is called Indefinite integration. It is also known as antidifferentiation. ...
Answer to: First make a substitution and then use integration by parts to evaluate the integral. Integral of (arcsin(ln x))/x dx. By signing up,...
So now it is in the format ∫u v dx we can proceed: Differentiate u: u' = x' = 1 Integrate v: ∫v dx = ∫cos(x) dx = sin(x) (see Integration Rules) Now we can put it together: Simplify and solve: x sin(x) − ∫sin(x) dx x sin(x) + cos(x) + C Done!So...
∫xndx=⎧⎪⎨⎪⎩log(x)xn+1n+1ifn=−1otherwise. int(x^n)orint(x^n,x) π/2∫0sin(2x)dx=1 int(sin(2*x), 0, pi/2)orint(sin(2*x), x, 0, pi/2) g= cos(at+b) ∫g(t)dt=sin(at+b)/a g = cos(a*t + b) int(g)orint(g, t) ...
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∫u v dx. example: ∫ x lnx dx by using the integration by parts method, we get here, the function is integrated with respect to x, then it becomes ∫ x lnx dx = (½)x 2 log x -(x 2 /4) + c comments leave a comment cancel reply your mobile number and email id will ...
(x)dx$(with periodicf(x))Improper integrals over (0, ∞)$\\int^{\\infty}_{0}f(x)dx$$\\int^{\\infty}_{0}f(x)(\\log x)^{p}dx$(p, positive integers)$\\int^{\\infty}_{0}x^{\\alpha}f(x)dx$(αε C)The Integral$\\int^{x_{0} + i\\infty}_{x_{0} - i\\...
du=ddx(2x)dx=2xlog(2)dx v=∫exdx=ex 3. Apply the Integration by Parts Formula: The integration by parts formula is: ∫udv=uv−∫vdu Substituting the values we have: I=2xex−∫ex(2xlog(2))dx 4. Simplify the Integral: The integral we have now is: I=2xex−log(2)∫2xexdx...
{\mathrm dx\over x}+1-\int_{1-\varepsilon}^N{\{x\}\over x^2}\mathrm dx \\ &=\int_1^N{\mathrm dx\over x}+1-\int_1^N{\{x\}\over x^2}\mathrm dx \\ &=\log N+1-\int_1^\infty{\{x\}\over x^2}\mathrm dx+\int_N^\infty{\{x\}\over x^2}\mathrm dx \\ ...