Answer to Question #208181 in Calculus for grusha

Question

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

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

Answer 1

$1) \int (5x^4 - 4x^3 +2x-3)\,dx =x^5 - x^4+x^2-3x +C \\ 2) \int (3\sin(x)-2\cos(x))\,dx =-3\cos(x)-2\sin(x) +C\\ 3) \int (x^3+x)\cos(5x)\,dx= \\ = (x^3+x)\frac{1}{5}\sin(5x)-\int\frac{1}{5}(3x^2+1)\sin(5x)\,dx =\\ = (x^3+x)\frac{1}{5}\sin(5x) + \frac{1}{25}(3x^2+1)\cos(5x)-\frac{1}{25}\int6x\cos(5x)\,dx=\\ =(x^3+x)\frac{1}{5}\sin(5x) + \frac{1}{25}(3x^2+1)\cos(5x)-\frac{1}{125}6x\sin(5x)+\\+\frac{6}{5}\int\sin(5x)\,dx=(x^3+x)\frac{1}{5}\sin(5x) + \frac{1}{25}(3x^2+1)\cos(5x)-\\ -\frac{1}{125}6x\sin(5x)-\frac{6}{625}\cos(5x)+C = \\ =(\frac{x^3}{5}+\frac{19x}{125})\sin(5x)+(\frac{3x^2}{25}+\frac{19}{625})\cos(5x)+C\\ 4)\int(e^{2x}+\sin(3x))\,dx=\frac{1}{2}e^{2x}-\frac{1}{3}\cos(3x)+C$