find integral using table of integrals )
integral sin^4xdx
this the formula i used
integral sin^n xdx =-1/n sin^n-1xcosx +n-1/n integral sin^n-2
using the formula this is what i got: integral sin^4xdx=-1/4sin^3xcosx+3/4 integral sin^2xdx= -1/2sinxcosx+1/2 integral 1 dx
can someone please show me how to combine the two integrals to get the final answer.
My answer, "|" = integral symbol
|sin^4 x dx =
-1/4 sin^3 x cos x + 3/4 (-1/2 sin x cos x + 1/2 | sin x dx )
-1/4 sin^3 x cos x + 3/4 (-1/2 sin x cos x + 1/2 (-cos x)) + C
-1/4 sin^3 x cos x + 3/4 (-1/2 sin x cos x - 1/2 cos x) + C
-1/4 sin^3 x cos x - 3/8 sin x cos x
- 3/8 cos x + C
Hold on a few minutes, I might have a mistake.
Mistake fixed below
-1/4 sin^3 x cos x + 3/4 (-1/2 sin x cos x + 1/2 | 1 dx )
-1/4 sin^3 x cos x + 3/4 (-1/2 sin x cos x + 1/2 x) + C
-1/4 sin^3 x cos x - 3/8 sin x cos x
+ 3/8 x + C
To combine the two integrals, start by examining the second integral:
∫(1)dx = ∫1dx = x + C
Next, plug this result back into the first integral:
-1/2∫sin(x)cos(x) dx + 1/2∫1 dx
To evaluate the first integral, use the following formula:
∫sin(x)cos(x) dx = -1/2 * sin^2(x) + C
Now, we can combine the two integrals:
-1/2 * (-1/2 * sin^2(x)) + 1/2 * x + C
Simplifying gives:
1/4 * sin^2(x) + 1/2 * x + C
So, the final answer is:
∫sin^4(x)dx = 1/4 * sin^2(x) + 1/2 * x + C