How do you integrate ∫sin2xdx?
If you want to find the integration of sin2x then you have to follow integration by substitutional together with the one of the known integral. Integration of sin2x can be written as ∫sin2x dx. As you know that sin2x can be written as sin(2x)=2sin(x)cos(x). So we just need to integrate 2sin(x)cos(x). We can rewrite as ∫2cos(x)sin(x)dx, after tanking 2 out it can be written as 2∫cos(x)sin(x)dx
Now we know that ∫sin(x)cos(x)dx = −cos2(x)/2+C. So 2∫cos(x)sin(x)dx = −cos2(x)+C. where C is a integration constant term.
What is the integration of sin2x
Now we will drive the formula of ∫sin2x dx by using integration by substitution method. We will assume that 2x = u such that d2x = du, so 2dx = du now we can say that dx = (1/2)du. Lets start the solution of the question –
- Consider ∫sin2x dx
- Let u = 2x such that du = 2dx ⇒ dx = (½)du.
- Substituting the above equations in ∫sin2x dx, we get;
- ∫ sin2x dx = ∫ sin u (½) du
- = (½) ∫ sin u du
- = (½) (-cos u) + C where C is a integration constant
- ∫sin2x dx = -(½) cos2x + C
- Therefore, ∫sin2x dx = -(½) cos2x + C
∫2x dx = x2 + C, with C as the integration constant
∫sin2x dx = -(½) cos2x + C, with C as the integration constant
∫sin 3x dx = (-1/3) cos 3x + C, where C is the constant of integration
The integral of cos(2x) is (1/2)sin(2x) + C, where C is a constant.
sin2x = 2 sin x cos x and sin2x = (2 tan x)/(1 + tan2x)
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