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

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

Go!
1
2
3
4
5
6
7
8
9
0
x
y
(◻)
◻/◻
2

e
π
ln
log
lim
d/dx
d/dx
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

Answer

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

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

Add the constant of integration

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

Answer

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

Struggling with math?

Access detailed step by step solutions to millions of problems, growing every day!

Problem Analysis

Main topic:

Integration by substitution

Time to solve it:

0.23 seconds

Views:

93