Mathematics Calculus Integral Calculus
how would you do the integral of the sin(square root x)
? I just can't seem to get started
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(integral) e^3x dx A. e^3x+C B. 1/3e^3x+C C. e^4x+C D. 1/4e^4x+C Evaluate (integral) dx/(Square root 9-8x-x^2) A. sin^-1(x+4/5)+C
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if u = 3x you have integral e^u du/3, so (B) since 9-8x-x^2 = 25 - (x+4)^2, and integral
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To find the derivative of the function, y = ∫√(x) to x^3 (√t * sin(t))dt, we can use the
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My answer, "|" = integral symbol |sin^4 x dx = -1/4 sin^3 x cos x + 3/4 (-1/2 sin x cos x + 1/2 |
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You can proceed as follows. Substitute: x = e^t. Then the integral becomes: Integral from ln(2) to
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well, d/dx √x = 1/(2√x) so what do you think?
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The integral is (-cos x + sin x)[0,π/6] = (-√3/2 + 1/2) - (-1 + 0) = 3/2 - √3/2 = 0.63397
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Does ^3square root mean cube root? I'll assume so. Roots are just fractional exponents, so y = 7x
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#2 and #3 are the same. #1: u = x-1 #4: u = x^2-1
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I could easily integrate #1 and #4 I see no difference isn #2 and #3, but would have difficulty with
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#2 and #2 look the same to me. note that d(x^2) = 2x dx
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