#math problem 14-07-2026 Calculus problem.∫∫∫ Compute this indefinite integral.

@MathMath901 ∫ sin θ d(cos θ) = ∫ (sin θ. (–sin θ dθ)) = ∫ (–sin² θ. dθ) = ∫ (–½ (1 – cos 2θ)) dθ = ∫ (–½ dθ) + ∫ (½ cos 2θ dθ) = –½ θ + ½. ½ sin 2θ + C = –½ θ + ¼ sin 2θ + C

@MathMath901 1. intergrate sin (theta) d(cos(theta)) start with d cos theta = - sin theta d theta and intergrate sin = sin^squared theta add 1 to the power divide by the old power sin^2/1 = sin^2 -- the solution for first intergral is this: sin^2(theta) * - sin(d (theta))

@MathMath901 acknowlegde the problem here are my steps: lookup table deriavation table : 1. sin -> 2. cos ->3. -sin -> 4. -cos rewrite the indefinite intergral step deriavation -1 from the power then mutiply by the old power intergration +1 to the power divide by the new power

@MathMath901 now for the third: 1/2 sin(theta) ^3 - cos(theta) +C = (sin(theta) ^ 3*1/2 = 1/2*3 =1/4 * 1/2 = 1/8 * - (-sin(theta))

@MathMath901 Hmmm ... isn't d(cos(\theta)) = -sin(\theta) d(\theta).

@MathMath901 now the double intergral sin^2 * -sin(d(theta)) ---- sin^3/2 = 1/2 sin(sin^3)* - cos(theta) + C

@MathMath901 i still got it wrong