# Integrate -1cos(3x)+7sin(4x)

## \int\left(\left(-1\right)\cdot\cos\left(3x\right)+7\sin\left(4x\right)\right)dx

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$-\frac{7}{4}\cos\left(4x\right)-\frac{1}{3}\sin\left(3x\right)+C_0$

## Step by step solution

Problem

$\int\left(\left(-1\right)\cdot\cos\left(3x\right)+7\sin\left(4x\right)\right)dx$
1

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

$\int7\sin\left(4x\right)dx+\int-\cos\left(3x\right)dx$
2

Taking the constant out of the integral

$7\int\sin\left(4x\right)dx+\int-\cos\left(3x\right)dx$
3

Taking the constant out of the integral

$7\int\sin\left(4x\right)dx-\int\cos\left(3x\right)dx$
4

Apply the formula: $\int\sin\left(x\cdot a\right)dx$$=-\frac{1}{a}\cos\left(x\cdot a\right), where a=4 7\cdot \frac{1}{4}\left(-1\right)\cos\left(4x\right)-\int\cos\left(3x\right)dx 5 Multiply \frac{7}{4} times -1 -\int\cos\left(3x\right)dx-\frac{7}{4}\cos\left(4x\right) 6 Apply the formula: \int\cos\left(x\cdot a\right)dx$$=\frac{1}{a}\sin\left(x\cdot a\right)$, where $a=3$

$-1\cdot \frac{1}{3}\sin\left(3x\right)-\frac{7}{4}\cos\left(4x\right)$
7

Multiply $\frac{1}{3}$ times $-1$

$-\frac{1}{3}\sin\left(3x\right)-\frac{7}{4}\cos\left(4x\right)$
8

$-\frac{7}{4}\cos\left(4x\right)-\frac{1}{3}\sin\left(3x\right)+C_0$

$-\frac{7}{4}\cos\left(4x\right)-\frac{1}{3}\sin\left(3x\right)+C_0$

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### Main topic:

Integration by substitution

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