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Power series Calculator

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1

Solved example of power series

$\int\sin\left(4x\right)\cdot\cos\left(2x\right)dx$
2

Apply the formula: $\sin\left(x\right)\cos\left(y\right)$$=\frac{\sin\left(x+y\right)+\sin\left(x-y\right)}{2}$, where $x=4x$ and $y=2x$

$\int\frac{\sin\left(4x+2x\right)+\sin\left(4x-2x\right)}{2}dx$

Adding $4x$ and $2x$

$\int\frac{\sin\left(6x\right)+\sin\left(4x-2x\right)}{2}dx$

Adding $4x$ and $-2x$

$\int\frac{\sin\left(6x\right)+\sin\left(2x\right)}{2}dx$
3

Simplifying

$\int\frac{\sin\left(6x\right)+\sin\left(2x\right)}{2}dx$
4

Split the fraction $\frac{\sin\left(6x\right)+\sin\left(2x\right)}{2}$ inside the integral, in two terms with common denominator $2$

$\int\left(\frac{\sin\left(6x\right)}{2}+\frac{\sin\left(2x\right)}{2}\right)dx$
5

The integral of the sum of two or more functions is equal to the sum of their integrals

$\int\frac{\sin\left(6x\right)}{2}dx+\int\frac{\sin\left(2x\right)}{2}dx$

Take the constant out of the integral

$\frac{1}{2}\int\sin\left(6x\right)dx$

Apply the formula: $\int\sin\left(ax\right)dx$$=-\left(\frac{1}{a}\right)\cos\left(ax\right)$, where $a=6$

$-\frac{1}{12}\cos\left(6x\right)$
6

The integral $\int\frac{\sin\left(6x\right)}{2}dx$ results in: $-\frac{1}{12}\cos\left(6x\right)$

$-\frac{1}{12}\cos\left(6x\right)$

Take the constant out of the integral

$\frac{1}{2}\int\sin\left(2x\right)dx$

Apply the formula: $\int\sin\left(ax\right)dx$$=-\left(\frac{1}{a}\right)\cos\left(ax\right)$, where $a=2$

$-\frac{1}{4}\cos\left(2x\right)$
7

The integral $\int\frac{\sin\left(2x\right)}{2}dx$ results in: $-\frac{1}{4}\cos\left(2x\right)$

$-\frac{1}{4}\cos\left(2x\right)$
8

Gather the results of all integrals

$-\frac{1}{12}\cos\left(6x\right)-\frac{1}{4}\cos\left(2x\right)$
9

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration

$-\frac{1}{12}\cos\left(6x\right)-\frac{1}{4}\cos\left(2x\right)+C_0$

Final Answer

$-\frac{1}{12}\cos\left(6x\right)-\frac{1}{4}\cos\left(2x\right)+C_0$

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