How do I solve this integrals problem?

[lasvegas666]

 

[mario99]

View attachment 37811

Integrate both sides.

 

[lasvegas666]

Integrate both sides.

Could you elaborate a little bit?

How do I integrate by both of the sides?

 

[blamocur]

Could you elaborate a little bit?

How do I integrate by both of the sides?

What have you tried so far? What do you know about integration?

 

[mario99]

Could you elaborate a little bit?

How do I integrate by both of the sides?

hint: [imath]\displaystyle f'(x) = \frac{dy}{dx}[/imath]

 

[lasvegas666]

hint: [imath]\displaystyle f'(x) = \frac{dy}{dx}[/imath]

So, -2sinx equals 2cosx.

I got the answer.

 

[mario99]

So, -2sinx equals 2cosx.

I got the answer.

Yes, and from the given condition,[imath]\displaystyle \ f\left(\frac{\pi}{2}\right) = 1\ [/imath], you can find the constant of integration.

 

[khansaheb]

So, -2sinx equals 2cosx.

That statement above is incorrect.

 

[NotWindowsExplorer]

That statement above is incorrect.

i think lasvegas666 meant to say "the antiderivative of -2sinx equals 2cosx"

 

[mario99]

i think lasvegas666 meant to say "the antiderivative of -2sinx equals 2cosx"

We knew he meant that. Professor Khan wanted him to say it literally or symbolically.

 

[Steven G]

So, -2sinx equals 2cosx.

I got the answer.

No! -2sinx does NOT equal 2cosx.
You need to say what you mean and use the equal sign ONLY when both sides are equal (or if you think that both sides are equal).
You mean to say that integral of -2sinx = 2cosx. And even that is wrong! integral of -2sinx = 2cosx + C
Now use the other piece of information you were given to solve for C

 

[Steven G]

We knew he meant that. Professor Khan wanted him to say it literally or symbolically.

Yes, we all knew what the OP said, but Khan wanted the OP to say it correctly.
I too posted the same information hoping to get it in the OP's head.