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

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

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

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

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

Solution steps

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

y = 1 0^{x}

(\frac{1}{3}) (\frac{d x}{d t}) - 2 e^{- t} = - x

\int_{0}^{1} x (1 - x)^{2} d x

x \to 0 lim (x (\frac{1}{x}))

x \to 8 lim (\frac{6}{x - 8})

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