Question #208181

integrate the following with respect to x up to a constant C:

  • 5x4- 4x3 + 2x - 3
  • 3sin x - 2cos x
  • (x3 + x)cos 5x
  • e2x + sin3x

Expert's answer

1)∫(5x4−4x3+2x−3) dx=x5−x4+x2−3x+C2)∫(3sin⁡(x)−2cos⁡(x)) dx=−3cos⁡(x)−2sin⁡(x)+C3)∫(x3+x)cos⁡(5x) dx==(x3+x)15sin⁡(5x)−∫15(3x2+1)sin⁡(5x) dx==(x3+x)15sin⁡(5x)+125(3x2+1)cos⁡(5x)−125∫6xcos⁡(5x) dx==(x3+x)15sin⁡(5x)+125(3x2+1)cos⁡(5x)−11256xsin⁡(5x)++65∫sin⁡(5x) dx=(x3+x)15sin⁡(5x)+125(3x2+1)cos⁡(5x)−−11256xsin⁡(5x)−6625cos⁡(5x)+C==(x35+19x125)sin⁡(5x)+(3x225+19625)cos⁡(5x)+C4)∫(e2x+sin⁡(3x)) dx=12e2x−13cos⁡(3x)+C1) \int (5x^4 - 4x^3 +2x-3)\,dx =x^5 - x^4+x^2-3x +C \\ 2) \int (3\sin(x)-2\cos(x))\,dx =-3\cos(x)-2\sin(x) +C\\ 3) \int (x^3+x)\cos(5x)\,dx= \\ = (x^3+x)\frac{1}{5}\sin(5x)-\int\frac{1}{5}(3x^2+1)\sin(5x)\,dx =\\ = (x^3+x)\frac{1}{5}\sin(5x) + \frac{1}{25}(3x^2+1)\cos(5x)-\frac{1}{25}\int6x\cos(5x)\,dx=\\ =(x^3+x)\frac{1}{5}\sin(5x) + \frac{1}{25}(3x^2+1)\cos(5x)-\frac{1}{125}6x\sin(5x)+\\+\frac{6}{5}\int\sin(5x)\,dx=(x^3+x)\frac{1}{5}\sin(5x) + \frac{1}{25}(3x^2+1)\cos(5x)-\\ -\frac{1}{125}6x\sin(5x)-\frac{6}{625}\cos(5x)+C = \\ =(\frac{x^3}{5}+\frac{19x}{125})\sin(5x)+(\frac{3x^2}{25}+\frac{19}{625})\cos(5x)+C\\ 4)\int(e^{2x}+\sin(3x))\,dx=\frac{1}{2}e^{2x}-\frac{1}{3}\cos(3x)+C


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