How do you use substitution to integrate $(x^{2}) ({sin x}^{3})$ $(x^2)(sinx^3)$ ?

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Answer 1 · 2015-07-27T21:50:11

$\int (x^{2}) ({sin x}^{3}) d x = - \frac{1}{3} {cos x}^{3} + C$ $int (x^2)(sinx^3) dx = -1/3cosx^3 +C$

$\int (x^{2}) ({sin x}^{3}) d x$ $int (x^2)(sinx^3) dx$

Let $u = x^{3}$ $u = x^3$ so that $d u = 3 x^{2} d x$ $du = 3x^2 dx$

Out integral becomes:

$\frac{1}{3} \int ({sin x}^{3}) 3 x^{2} d x = \frac{1}{3} \int sin u d u$ $1/3 int (sinx^3) 3x^2dx = 1/3 int sin u du$

$= \frac{1}{3} (- cos u) + C$ $= 1/3 (-cosu) +C$

$= - \frac{1}{3} {cos x}^{3} + C$ $= -1/3cosx^3 +C$

Check the answer by differentiating:

$\frac{d}{d x} (- \frac{1}{3} {cos x}^{3} + C) = - \frac{1}{3} \cdot - sin (x^{3}) \cdot 3 x^{2}$ $d/dx(-1/3cosx^3 +C) = -1/3* -sin(x^3) * 3x^2$

$= sin (x^{3}) \cdot x^{2}$ $= sin(x^3)* x^2$

Looks good, so that's it.

How do you use substitution to integrate (x^2)(sinx^3) (x2)(sinx3)(x^2)(sinx^3) ?