Question

∫cos4xdx

calculate the indefinite integral

Answer 1

We have to find,

$\displaystyle\int\cos(4x)dx$

Let,

$u=4x\\\\\\\dfrac {du}{dx}=4\quad\implies\quad dx=\dfrac{du}{4}$

Now,

$\displaystyle\int\cos(4x)dx=\int\cos u\ \dfrac {du}{4}\\\\\\=\dfrac {1}{4}\int\cos u\ du=\dfrac {1}{4}\sin u+c\\\\\\=\mathbf{\dfrac{\sin(4x)}{4}+c}$

where $c$ is the integral constant.

Answer 2

Answer:

$\bold</strong><strong>\</strong><strong>r</strong><strong>e</strong><strong>d</strong><strong>{\frac{sin \: 4m}{4} + c}$

Step-by-step explanation:

we have to integrate,

$\int \: cos4x \: dx$

Let,

$4x = m$

Differentiate both sides wrt x,

we get,

$= > 4 \: dx = dm \\ \\ = > dx = \frac{dm}{4}$

Putting the value of dx,

we get,

$\int \frac{1}{4} \: cos m \: dm \\ \\ = \frac{1}{4} \int \: cosm \: dm$

But,

we know that,

$\int \cos( \alpha ) = \sin( \alpha )$

Therefore,

we get,

$= \frac{sin \: m}{4} + c$

So, putting the value of 'm'

we get,

Integration

$\bold{\frac{sin \: 4m}{4} + c}$

where,

C is an arbitrary constant