What is the integral of sin^4(r)cos^3(r) ?

The integral of sin^4(r)cos^3(r) is (sin^5(r))/5-(sin^7(r))/7+C

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integral of sin^4(r)cos^3(r)

\int sin^{4} (r) cos^{3} (r) d r

\frac{sin ^{5} ( r )}{5} - \frac{sin ^{7} ( r )}{7} + C

Solution steps

\int sin^{4} (r) cos^{3} (r) d r

Rewrite using trig identities

= \int (1 - sin^{2} (r)) cos (r) sin^{4} (r) d r

Apply u-substitution

: \int u^{4} (1 - u^{2}) d u

= \int u^{4} (1 - u^{2}) d u

Expand

u^{4} (1 - u^{2}) : u^{4} - u^{6}

= \int u^{4} - u^{6} 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^{4} d u - \int u^{6} d u

\int u^{4} d u = \frac{u ^{5}}{5}

\int u^{6} d u = \frac{u ^{7}}{7}

= \frac{u ^{5}}{5} - \frac{u ^{7}}{7}

Substitute back

u = sin (r)

= \frac{sin ^{5} ( r )}{5} - \frac{sin ^{7} ( r )}{7}

Add a constant to the solution

= \frac{sin ^{5} ( r )}{5} - \frac{sin ^{7} ( r )}{7} + C

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What is the integral of sin^4(r)cos^3(r) ?
The integral of sin^4(r)cos^3(r) is (sin^5(r))/5-(sin^7(r))/7+C