what is integral (sin 5x)dx ?

∫(sin 5x)dx = ∫(sin 5x)/5 d(5x) = -cos(5x)/5 + c

Integral (sin5x) dx

Rule of thumb: Every time you're taking the integral of a function with its insides being linear (in this case, 5x is linear since x is a power of 1), it is easy to do mental substitution.

All you have to do is take the integral of sin(x) mentally (the integral of sinx is -cosx), and apply it with the 5x (so we would get -cos5x). However, when you take the derivative of -cos(5x), you would get 5sin(5x), because the chain rule would force you to multiply 5. To offset this, you merely multiply by (1/5) to the integral. Therefore

Integral (sin(5x))dx = (-1/5)cos(5x) + C

If you'd like to actually see the substitution being done, I'll show you with u substitution.

Integral (sin(5x))dx

Let u = 5x

du = 5 dx

(1/5) du = dx

Therefore,

Integral (sin(5x))dx = Integral (sin(u) (1/5)du)

Pulling out the constants out of the integral, we get

(1/5) * Integral (sin(u)du)

Integrating appropriately, we have

(1/5) [-cos(u)] + C

(-1/5) cos(u) + C

Replacing u = 5x,

(-1/5) cos(5x) + C

∫(sin 5x)dx = ∫(sin 5x)/5 d(5x) = -cos(5x)/5 + c

Here, I used mental substitution. If you don't like it, you can substitute u = 5x. Either way, you should get the same answer.

(-(cos 5x)/5)+c

-(cos5x)/5 +C. if it was 'a' instead of 5, then it would have been -(cosax)/a +C

it can be solved by the reduction formula