why $\int{\cos({\pi}t)} dt = \frac{1}{\pi}\sin({\pi}t)$?

Question

Why?

$$\int{\cos({\pi}t)} dt = \frac{1}{\pi}\sin({\pi}t)$$

the indefinite integral of $ \cos(x) = \sin(x) $ isn't it?

so this suppose to be $$\int{\cos({\pi}t)} dt = \sin({\pi}t)$$

Why in the video at the minute 3:50 (https://www.youtube.com/watch?v=EuWaisQj3hU&index=29&list=PLF5E22224459D23D9) the professor says the first one?

Am I missing something?

please explain to me step by step and in detail. Thank you very much.

Answer 1

Use $u$-substitution, let $u = \pi t$, then $$\int \cos(\pi t)\,dt = \int \cos(u)\cdot\frac{1}{\pi}\,du = \frac{1}{\pi}\sin(\pi t)+C$$

To check your work, you can take the derivative of the RHS to check that it matches the integrand: $$\left(\frac{1}{\pi}\sin(\pi t)+C\right)' = \frac{1}{\pi}\cos (\pi t)\cdot \pi+0 = \cos(\pi t),$$ which matches the original integrand.

In fact, he explicitly says to take the derivative of "something" and that should match the integrand. That "something" is $$\frac{1}{\pi}\sin(\pi t)+C.$$ Remember that the derivative of a constant is zero.

Answer 2

Since you are asking why:

Due to chain rule, when we differentiate $\sin\pi t$ wrt $t$, we get an additional constant $\pi$ which is the coefficient of $t$

$$\dfrac{d(\sin\pi t)}{dt} = \cos\pi t \cdot \pi$$

So when we differentiate $\dfrac{\sin\pi t}{\pi}$ wrt $t$, we get $\cos\pi$

$$\dfrac{d}{dt}\left(\frac{\sin\pi t}{\pi}\right) = \cos\pi t $$

hence $$\int \cos(\pi t) \ dt = \frac{\sin\pi t}{\pi} + C$$

Answer 3

$\int \cos \left(\pi x\right)dx$

$\mathrm{Apply\:Integral\:Substitution:}\:\int > f\left(g\left(x\right)\right)\cdot g^{'}\left(x\right)dx=\int > f\left(u\right)du,\:\quad u=g\left(x\right)$ $u=\pi x\quad \:du=\pi > dx$

$\mathrm{Substitute:}\:u=\pi x$

$\frac{du}{dx}=\pi $

$\frac{d}{dx}\left(\pi x\right)$

$\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot > f\right)^{'}=a\cdot f^{'}$

$=\pi \frac{d}{dx}\left(x\right)$

$\mathrm{Apply\:the\:common\:derivative}:\quad > \frac{d}{dx}\left(x\right)=1$

$=\pi 1$

$\mathrm{Simplify}$

$=\pi $

$\Rightarrow \:du=\pi dx$

$\Rightarrow \:dx=\frac{1}{\pi }du$

$=\int \cos \left(u\right)\frac{1}{\pi }du$

$=\int \frac{\cos \left(u\right)}{\pi }du$

$=\int \frac{\cos \left(u\right)}{\pi }du$

$=\int \frac{\cos \left(u\right)}{\pi }du$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot > f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=\frac{1}{\pi }\int \cos \left(u\right)du$

$\mathrm{Use\:the\:common\:integral}:\quad \int \:\cos > \left(u\right)du=\sin \left(u\right)$

$=\frac{1}{\pi }\sin \left(u\right)$

$\mathrm{Substitute\:back}\:u=\pi x$

$=\frac{1}{\pi }\sin \left(\pi x\right)$

$\mathrm{Simplify}$

$=\frac{\sin \left(\pi x\right)}{\pi }$

$Add\:a\:constant\:to\:the\:solution$

$=\frac{\sin \left(\pi x\right)}{\pi }+C$

Answer 4

The derivative $f'(x)$ is the slope of the tangent at point $x$ to the curve $y=f(x)$. If you change $x$ into $\pi x$, you are shrinking the curve : for instance $\sin(\pi x)$ oscillates $\pi$ times faster that $\sin(x)$. This is why the factor $\pi$ is necessary : the slope of the tangent is $\pi$ times larger.