What is the integral of sin^3(x)cos^3(x) ?

The integral of sin^3(x)cos^3(x) is -(cos^4(x))/4+(cos^6(x))/6+C

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integral of sin^3(x)cos^3(x)

integral

sin^{3} (x) cos^{3} (x)

- \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{6} + C

Solution steps

\int sin^{3} (x) cos^{3} (x) d x

Rewrite using trig identities

= \int (1 - cos^{2} (x)) sin (x) cos^{3} (x) d x

Apply u-substitution

: \int - u^{3} (1 - u^{2}) d u

= \int - u^{3} (1 - u^{2}) d u

Expand

- u^{3} (1 - u^{2}) : - u^{3} + u^{5}

= \int - u^{3} + u^{5} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int u^{3} d u + \int u^{5} d u

\int u^{3} d u = \frac{u ^{4}}{4}

\int u^{5} d u = \frac{u ^{6}}{6}

= - \frac{u ^{4}}{4} + \frac{u ^{6}}{6}

Substitute back

u = cos (x)

= - \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{6}

Add a constant to the solution

= - \frac{cos ^{4} ( x )}{4} + \frac{cos ^{6} ( x )}{6} + C

(\frac{( x - 1 )}{( x + 1 )})^{\frac{1}{2}}

sin^{2} (x) cos^{3} (x)

\frac{sin ( x )}{( 1 - sin ( x ) )}

\frac{1}{v ^{4}}

\frac{( x ^{2} - 1 )}{( x ^{2} - 4 x + 1 )}

What is the integral of sin^3(x)cos^3(x) ?
The integral of sin^3(x)cos^3(x) is -(cos^4(x))/4+(cos^6(x))/6+C