find indefinite integral [(sinx)^3]*[(cosx)^3] dx
i got 1/192 * ((cos(6x)-9cos(2x)) + C
but this isnt right. explain plz? thnx
To find the indefinite integral of the function [(sin(x))^3]*[(cos(x))^3] dx, we can use the technique of u-substitution.
Let's start by making a substitution:
u = sin(x)
Now, we need to find the derivative of u with respect to x in order to convert the expression from dx to du.
du = cos(x) dx
Next, we need to express the entire expression in terms of u instead of x. We can do this by using the trigonometric identity: sin^2(x) + cos^2(x) = 1.
Rearranging the identity, we get: cos^2(x) = 1 - sin^2(x)
Now, let's rewrite the original expression using this identity:
[(sin(x))^3]*[(cos(x))^3] = (sin(x))^3 * [(cos(x))^2] * cos(x)
= (sin(x))^3 * (1 - sin^2(x)) * cos(x)
Substituting u = sin(x) and du = cos(x) dx, the integral becomes:
∫ (sin(x))^3 * (1 - sin^2(x)) * cos(x) dx = ∫ u^3 * (1 - u^2) du
Expanding the expression inside the integral:
= ∫ (u^3 - u^5) du
= ∫ u^3 du - ∫ u^5 du
Integrating term by term:
= (u^4/4) - (u^6/6) + C
Replacing u with sin(x) gives us:
= (sin^4(x)/4) - (sin^6(x)/6) + C
Therefore, the correct indefinite integral of [(sin(x))^3]*[(cos(x))^3] dx is:
∫ [(sin(x))^3]*[(cos(x))^3] dx = (sin^4(x)/4) - (sin^6(x)/6) + C
Your previous answer, 1/192 * (cos(6x) - 9cos(2x)) + C, is incorrect. It appears that there was an error made during the integration process.