What is the integral of cos^9(x)tan^8(x) ?

The integral of cos^9(x)tan^8(x) is (sin^9(x))/9+C

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integral of cos^9(x)tan^8(x)

\int cos^{9} (x) tan^{8} (x) d x

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

Solution steps

\int cos^{9} (x) tan^{8} (x) d x

Rewrite using trig identities

= \int sin^{8} (x) cos (x) d x

Apply u-substitution

: \int u^{8} d u

= \int u^{8} d u

Apply the Power Rule

: \frac{u ^{9}}{9}

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

Substitute back

u = sin (x)

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (3 e^{x})

f (2) = 11 x^{2} + 5 x

f (x) = x^{2} - x^{4}, (1, 0)

\frac{d}{d x} ((\frac{x + 2}{x})^{3} + (\frac{x - 2}{x})^{3})

\frac{d x}{d y} - \frac{x}{y} = 2 y

What is the integral of cos^9(x)tan^8(x) ?
The integral of cos^9(x)tan^8(x) is (sin^9(x))/9+C