int cosxsinx dx
integral cos(x) sin(x) dx
We simply use substitution.
Let u = sin(x)
Thus du = cos(x) dx
Rewriting,
= integral (u du)
= (1/2)*u^2 + C
= (1/2)*sin^2 (x) + C
Or you may also use substitute the formula sin(2x) = 2sin(x)cos(x) to the original, and then directly integrate.
Hope this helps~ :3
Or
following Jai's method.
let u = cosx
etc
to get
(-1/2) cos^2 x + c
To solve the integral of cos(x)sin(x)dx, you can use a technique called integration by parts. The formula for integration by parts is:
∫ u * dv = u * v - ∫ v * du
Let's assign u = cos(x) and dv = sin(x)dx. To find du and v, we can differentiate u and integrate dv respectively:
du = -sin(x)dx (differentiate cos(x))
v = -cos(x) (integrate sin(x)dx)
Now, we can use the integration by parts formula:
∫ cos(x)sin(x)dx = u * v - ∫ v * du
Substituting the values we found:
∫ cos(x)sin(x)dx = cos(x) * (-cos(x)) - ∫ (-cos(x)) * (-sin(x))dx
= -cos^2(x) + ∫ cos(x)sin(x)dx
At this point, we have a new integral on the right side, but it is the same as the original integral. So, we can substitute the integral on the right side with "I":
∫ cos(x)sin(x)dx = -cos^2(x) + I
Now we can solve for I:
∫ cos(x)sin(x)dx = -cos^2(x) + I
To find I, we can rearrange the equation:
I = ∫ cos(x)sin(x)dx + cos^2(x)
Now, we have two integrals: ∫ cos(x)sin(x)dx and ∫ cos^2(x)dx.
∫ cos(x)sin(x)dx can be solved using integration by parts again:
Let u = sin(x) and dv = cos(x)dx
Then, du = cos(x)dx and v = sin(x)
∫ cos(x)sin(x)dx = sin(x) * sin(x) - ∫ sin(x) * cos(x)dx
= sin^2(x) - ∫ sin(x) * cos(x)dx
Now, we have another integral on the right side. This integral is similar to the one we started with, but with the roles of sin(x) and cos(x) reversed. So let's assign it a new variable, J:
J = ∫ sin(x)cos(x)dx
Using the same technique, integration by parts, for J, let's assign u = cos(x) and dv = sin(x)dx:
Then, du = -sin(x)dx and v = -cos(x)
J = -cos(x) * cos(x) - ∫ (-cos(x)) * (-sin(x))dx
= -cos^2(x) + ∫ cos(x)sin(x)dx
We can substitute the integral we found for J back into the equation for ∫ cos(x)sin(x)dx:
J = -cos^2(x) + ∫ cos(x)sin(x)dx
Rearranging this equation, we get:
2J = -cos^2(x) + ∫ cos(x)sin(x)dx
Now, divide both sides by 2:
J = -1/2 * cos^2(x) + 1/2 * ∫ cos(x)sin(x)dx
Substituting J back into the equation for I:
I = ∫ cos(x)sin(x)dx + cos^2(x)
= -1/2 * cos^2(x) + 1/2 * ∫ cos(x)sin(x)dx + cos^2(x)
Simplifying:
I = -1/2 * cos^2(x) + 1/2 * ∫ cos(x)sin(x)dx + cos^2(x)
= -1/2 * cos^2(x) + 1/2 * ∫ cos(x)sin(x)dx + 1 * cos^2(x)
Now, we have two terms on the right side: -1/2 * cos^2(x) and 1 * cos^2(x). We can combine them:
I = -1/2 * cos^2(x) + 1/2 * cos^2(x) + ∫ cos(x)sin(x)dx
= 1/2 * ∫ cos(x)sin(x)dx
Therefore, the integral of cos(x)sin(x)dx is equal to (1/2) * ∫ cos(x)sin(x)dx.