How do you integrate ∫sin⁡2xdx

Question

vrangett · Accepted Answer

Using integration by substitution together with the known integral  sin⁡(x)dx=12∫sin⁡(2x)2dx  =12∫sin⁡(u)du  =12(−cos⁡(u))+C  =−12cos⁡(2x)+C

Fasaniu · Accepted Answer

So, now we have to integrate sin 2x ∫sin⁡2x=12∈2×sin⁡(2x)dx (1) Let us assume u=2x. Then du=2dx We know that ∫sin⁡x=−cos⁡x+C Hence on substituting, equation (1) becomes ∫sin⁡2xdx=12∫sin⁡(u)du ∫sin⁡2xdx=12(−cos⁡udu)+C ∫sin⁡2xdx=−12cos⁡(2x)+C

RizerMix · Accepted Answer

Since d is constant with respect to x, move d out of the integral.
$d \int \sin (2 x) x d x$
Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = x$ and $d v = \sin (2 x)$
$d (x (- \frac{1}{2} \cos (2 x)) - \int - \frac{1}{2} \cos (2 x) d x)$
Simplify,
$d (- \frac{x \cos (2 x)}{2} - \int - \frac{\cos (2 x)}{2} d x)$
Since -1 is constant with respect to x, move -1 out of the integral.
$d (- \frac{x \cos (2 x)}{2} - - \int \frac{\cos (2 x)}{2} d x)$
Simplify,
$d (- \frac{x \cos (2 x)}{2} + \int \frac{\cos (2 x)}{2} d x)$
Since $\frac{1}{2}$ is constant with respect to x, move $\frac{1}{2}$ out of the integral.
$d (- \frac{x \cos (2 x)}{2} + \frac{1}{2} \int \cos (u) \frac{1}{2} d u)$
Let $u = 2 x$ . Then $d u = 2 d x$ , so $\frac{1}{2} d u = d x$ . Rewrite using u and du.
$d (- \frac{x \cos (2 x)}{2} + \frac{1}{2} \int \cos (u) \frac{1}{2} d u)$
Combine $\cos (u)$ and $\frac{1}{2}$ .
$d (- \frac{x \cos (2 x)}{2} + \frac{1}{2} \int \frac{\cos (u)}{2} d u)$
Since $\frac{1}{2}$ is constant with respect to u, move $\frac{1}{2}$ out of the integral.
$d (- \frac{x \cos (2 x)}{2} + \frac{1}{2} (\frac{1}{2} \int \cos (u) d u))$
Simplify,
$d (- \frac{x \cos (2 x)}{2} + \frac{1}{4} \int \cos (u) d u)$
The integral of $\cos (u)$ with respect to u is $\sin (u)$
$d (- \frac{x \cos (2 x)}{2} + \frac{1}{4} (\sin (u) + C))$

How do you integrate \int\sin2xdx

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