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How would you solve the integral of 1 plus tan2x plus tan squared 2x?
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
What is the indefinte integral of sin2x?
The indefinite integral of sin 2x is -cos 2x / 2 + C, where C is any constant.
What is the integral of sin x Times Square x times exp x?
Assume the expression is: ∫ sin(x)x²e^x dx Then: Take the integral: integral e^x x^2 sin(x) dx For the integrand e^x x^2 sin(x), integrate by parts, integral f dg = f g- integral g df, where f = x^2, dg = e^x sin(x) dx, df = 2 x dx, g = 1/2 e^x (sin(x)-cos(x)): = 1/2 e^x x^2 sin(x)-1/2 (e^x x^2 cos(x))- integral e^x x (sin(x)-cos(x)) dx Expanding the integrand e^x x (sin(x)-cos(x)) gives e^x x sin(x)-e^x x cos(x): = 1/2 e^x x^2 sin(x)-1/2 (e^x x^2 cos(x))- integral (e^x x sin(x)-e^x x cos(x)) dx Integrate the sum term by term and factor out constants: = 1/2 e^x x^2 sin(x)-1/2 (e^x x^2 cos(x))- integral e^x x sin(x) dx+ integral e^x x cos(x) dx For the integrand e^x x sin(x), integrate by parts, integral f dg = f g- integral g df, where f = x, dg = e^x sin(x) dx, df = dx, g = 1/2 e^x (sin(x)-cos(x)): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x x sin(x)+1/2 e^x x cos(x)+ integral e^x x cos(x) dx+1/2 integral e^x (sin(x)-cos(x)) dx Expanding the integrand e^x (sin(x)-cos(x)) gives e^x sin(x)-e^x cos(x): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x x sin(x)+1/2 e^x x cos(x)+ integral e^x x cos(x) dx+1/2 integral (e^x sin(x)-e^x cos(x)) dx Integrate the sum term by term and factor out constants: = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x x sin(x)+1/2 e^x x cos(x)+1/2 integral e^x sin(x) dx-1/2 integral e^x cos(x) dx+ integral e^x x cos(x) dx For the integrand e^x cos(x), use the formula integral exp(alpha x) cos(beta x) dx = (exp(alpha x) (alpha cos(beta x)+beta sin(beta x)))/(alpha^2+beta^2): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/4 e^x sin(x)-1/2 e^x x sin(x)-1/4 (e^x cos(x))+1/2 e^x x cos(x)+1/2 integral e^x sin(x) dx+ integral e^x x cos(x) dx For the integrand e^x sin(x), use the formula integral exp(alpha x) sin(beta x) dx = (exp(alpha x) (alpha sin(beta x)-beta cos(beta x)))/(alpha^2+beta^2): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x x sin(x)-1/2 (e^x cos(x))+1/2 e^x x cos(x)+ integral e^x x cos(x) dx For the integrand e^x x cos(x), integrate by parts, integral f dg = f g- integral g df, where f = x, dg = e^x cos(x) dx, df = dx, g = 1/2 e^x (sin(x)+cos(x)): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)+e^x x cos(x)-1/2 e^x cos(x)-1/2 integral e^x (sin(x)+cos(x)) dx Expanding the integrand e^x (sin(x)+cos(x)) gives e^x sin(x)+e^x cos(x): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)+e^x x cos(x)-1/2 e^x cos(x)-1/2 integral (e^x sin(x)+e^x cos(x)) dx Integrate the sum term by term: = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)+e^x x cos(x)-1/2 e^x cos(x)-1/2 integral e^x sin(x) dx-1/2 integral e^x cos(x) dx For the integrand e^x cos(x), use the formula integral exp(alpha x) cos(beta x) dx = (exp(alpha x) (alpha cos(beta x)+beta sin(beta x)))/(alpha^2+beta^2): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/4 e^x sin(x)+e^x x cos(x)+-3/4 e^x cos(x)-1/2 integral e^x sin(x) dx For the integrand e^x sin(x), use the formula integral exp(alpha x) sin(beta x) dx = (exp(alpha x) (alpha sin(beta x)-beta cos(beta x)))/(alpha^2+beta^2): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x sin(x)+e^x x cos(x)-1/2 e^x cos(x)+constant Which is equal to: Answer: | | = 1/2 e^x ((x^2-1) sin(x)-(x-1)^2 cos(x))+constant
Int sin 3x cos 5x dx?
integral sin(3 x) cos(5 x) dx = 1/16 (8 cos^2(x)-cos(8 x))+C
What is cos x2?
cos(x^2)=cos(x times x)
