i can't fiugre how the solution is this...? can anyone explain me how?

Question

Find the antiderivative of
f(x) = sin3xdx
= -1/3 cos3x+c ...i know derivative sin is -cos which is
- cos 3x+c ..but where is the 1/3 from?

Also i have another problem that i got stucked!
h'(x)= (5x^2-7x) / (³√x) how is this done?

Answer 1

Apply the chain rule
y=cos z with z = 3x
dy/dx = dy/dz*dz/dx =-sinz* 3

Answer 2

I am going to put the word integral in place of the squiggly line.

Integral: Sin3xdx---Do u-substitution-->u=3x-->(f)'x = 3dx-->(now divide 3 for both sides, this is where you get the 1/3) 1/3du=dx...now put it into your integral

Integral: Sinu1/3du --> multiply 1/3 by the integral--> 1/3Integral: Sinudu

Find the anti-derivative of sinu and multiply 1/3 by what you get

1/3 * -cosu -->-1/3cosu-->(substitute u back) and you get..

-1/3Cos3x + C

For the second one:

(5x^2 -7x) * 1/(x)^-1/3 (I pulled out the 5x^2-7x) ---> then it's (5x^2 - 7x)(x^-1/3) -->multiply x^-1/3 in and you get-->(5x^5/3 - 7x^2/3) = (now find the anti-derivative of each) and you get..

15/8x^8/3 - 21/5x^5/3

Answer 3

antiderivative of 1/3 is 1/3x pretty logical? antiderivative of
-1/3 is -1/3x But I haven't been able to do calculus problems just yet, so I can't help out here

Answer 4

Derivative of sin(x) is -cos(x)
Derivative of sin(ax) is -acos(ax)
So ...
derivative of sin(3x) is -3cos(3x)
Anti derivative of sin(3x) is -(1/3)cos(3x)