What is the integral of tan^8(x)sec^2(x) ?

The integral of tan^8(x)sec^2(x) is (tan^9(x))/9+C

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integral of tan^8(x)sec^2(x)

\int tan^{8} (x) sec^{2} (x) d x

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

Solution steps

\int tan^{8} (x) sec^{2} (x) d x

Rewrite using trig identities

= \int (1 + tan^{2} (x)) tan^{8} (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 = tan (x)

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

Add a constant to the solution

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

x y^{^{'}} = - y - x, y (1) = 1

x \to - 2 lim (\frac{- x}{- x + 2})

(7 x)^{x}

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\int ((1 - \frac{2}{3} x^{2})^{6}) (x) d x

What is the integral of tan^8(x)sec^2(x) ?
The integral of tan^8(x)sec^2(x) is (tan^9(x))/9+C