Question

Find the integral of the given function w.r.t x

$y=sin\left(8x\right)+x$

• A. $-\frac{cos8x}{8}+\frac{x^2}{2}+c$
• B. $\frac{8cos8x}{8}+1+c$
• C. $\frac{cos8x}{8}+\frac{x^2}{2}+c$
• D. $\frac{cos8x}{8}+1+c$

Answer 1

The correct option is A

$-\frac{cos8x}{8}+\frac{x^2}{2}+c$

$\int \left(sin\right(8x)+x)dx=\int sin\left(8x\right)dx+\int \left(x\right)dx$

8x = t

8dx = dt

dx = $\frac{dt}{8}$

$\Rightarrow \int \frac{sintdt}{8}+\frac{x^2}{2}+c_1$ $(\because \int x^{n}dx=\frac{x^{n+1}}{n+1})+c$

$\Rightarrow -\frac{cost}{8}+c_o+\frac{x^2}{2}+c$ $(\because \int sintdt=-cost)$

$-\frac{cos8x}{8}+\frac{x^2}{2}+c$ $(c=c_0+c_1)$