What is the integral of (sin(x))^6(cos(x))^3 ?

The integral of (sin(x))^6(cos(x))^3 is (sin^7(x))/7-(sin^9(x))/9+C

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integral of (sin(x))^6(cos(x))^3

\int (sin (x))^{6} (cos (x))^{3} d x

\frac{sin ^{7} ( x )}{7} - \frac{sin ^{9} ( x )}{9} + C

Solution steps

\int (sin (x))^{6} (cos (x))^{3} d x

Simplify

= \int sin^{6} (x) cos^{3} (x) d x

Rewrite using trig identities

= \int (1 - sin^{2} (x)) cos (x) sin^{6} (x) d x

Apply u-substitution

: \int u^{6} (1 - u^{2}) d u

= \int u^{6} (1 - u^{2}) d u

Expand

u^{6} (1 - u^{2}) : u^{6} - u^{8}

= \int u^{6} - u^{8} 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^{6} d u - \int u^{8} d u

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

\int u^{8} d u = \frac{u ^{9}}{9}

= \frac{u ^{7}}{7} - \frac{u ^{9}}{9}

Substitute back

u = sin (x)

= \frac{sin ^{7} ( x )}{7} - \frac{sin ^{9} ( x )}{9}

Add a constant to the solution

= \frac{sin ^{7} ( x )}{7} - \frac{sin ^{9} ( x )}{9} + C

\int sec^{6} (4 x) d x

f (x) = (3 x^{3} - 4 x^{2} + 5 x - 9)^{2}, at x = 2

x \to 2 lim (x^{2} + 6 x + 5)

x \to 0 + lim (x^{13 x})

x \to 4 lim (6)

What is the integral of (sin(x))^6(cos(x))^3 ?
The integral of (sin(x))^6(cos(x))^3 is (sin^7(x))/7-(sin^9(x))/9+C