What is the integral of $10 sin (\frac{x}{5})$ $10sin(x/5)$ ?

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Answer 1 · 2015-04-22T14:53:18

It is $- 50 cos (\frac{x}{5})$ $-50cos(x/5)$

$\int 10 sin (\frac{x}{5}) d x = 10 \int sin (\frac{x}{5}) d x$ $int 10 sin(x/5) dx = 10 intsin(x/5) dx$

Let $u = \frac{x}{5}$ $u = x/5$ so that $d u = \frac{1}{5} d x$ $du = 1/5 dx$ and therefore $d x = 5 d u$ $dx = 5 du$

Substitute:

$10 \int sin (\frac{x}{5}) d x = 10 \int (sin u) 5 d u = 50 \int sin u d u$ $10 int sin(x/5) dx = 10 int (sinu) 5 du = 50 int sinu du$

Now integrate:

#10 int sin(x/5) dx = 50 (-cos u) +C
And finally, reverse the substitution:

$10 \int sin (\frac{x}{5}) d x = - 50 cos (\frac{x}{5}) + C$ $10 int sin(x/5) dx = -50 cos (x/5)+C$

What is the integral of 10sin(x5)10sin(x/5)?