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

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

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integral of sec^2(θ)tan^8(θ)

\int sec^{2} (θ) tan^{8} (θ) d θ

\frac{tan ^{9} ( θ )}{9} + C

Solution steps

\int sec^{2} (θ) tan^{8} (θ) d θ

Rewrite using trig identities

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

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 (θ)

= \frac{tan ^{9} ( θ )}{9}

Add a constant to the solution

= \frac{tan ^{9} ( θ )}{9} + C

\frac{\partial}{\partial x} (1 + y^{2})

9 y^{^{''}} - 6 y^{^{'}} + y = 0

n = 1 \sum \infty (\frac{1}{7})^{n}

4 x - x^{2}, x^{2}, 2, 0

x \to - 6 - lim (\frac{1}{( x + 6 ) ^{6}})

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