What is the integral of $\int {sin (x)}^{3} \cdot cos (x) d x$ $int sin(x)^3 * cos(x) dx$ ?

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Answer 1 · 2018-05-05T09:32:03

$= \frac{{sin}^{4} (x)}{4} + C$ $=(sin^4(x))/(4)+C$

$\int$ $int_$ ${sin}^{3} (x) \cdot cos (x) d x$ $sin^3(x)*cos(x) dx$

We can use substitution to remove $cos (x)$ $cos(x)$ . So, let's use $sin (x)$ $sin(x)$ as our source.

$u = sin (x)$ $u=sin(x)$

Which then means that we will get,

$\frac{d u}{d x} = cos (x)$ $(du)/(dx)=cos(x)$

Finding $d x$ $dx$ will give,

$d x = \frac{1}{cos (x)} \cdot d u$ $dx= 1/cos(x) * du$

Now replacing the original integral with the substitution,

$\int$ $int_$ $u^{3} \cdot cos (x) \cdot \frac{1}{cos (x)} d u$ $u^3*cos(x) * 1/cos(x) du$

We can cancel out $cos (x)$ $cos(x)$ here,

$\int$ $int_$ $u^{3} d u$ $u^3 du$

$= \frac{1}{3 + 1} u^{3 + 1} + C = \frac{1}{4} u^{4} + C$ $= 1/(3+1)u^(3+1)+C = 1/4 u^4+C$

Now setting in for $u$ $u$ ,

$= \frac{{sin (x)}^{4}}{4} + C = \frac{{sin}^{4} (x)}{4} + C$ $=sin(x)^4/4+C = sin^4(x)/4 + C$

What is the integral of ∫sin(x)3⋅cos(x)dxint sin(x)^3 * cos(x) dx?