What is the integral from 0 to pi/4 of tan^8(θ)sec^2(θ) ?

The integral from 0 to pi/4 of tan^8(θ)sec^2(θ) is 1/9

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integral from 0 to pi/4 of tan^8(θ)sec^2(θ)

\int_{0}^{\frac{π}{4}} tan^{8} (θ) sec^{2} (θ) d θ

\frac{1}{9}

Decimal

0.11111 \dots

Solution steps

\int_{0}^{\frac{π}{4}} tan^{8} (θ) sec^{2} (θ) d θ

Rewrite using trig identities

= \int_{0}^{\frac{π}{4}} (1 + tan^{2} (θ)) tan^{8} (θ) d θ

Apply u-substitution

= \int_{0}^{1} u^{8} d u

Apply the Power Rule

= [\frac{u ^{9}}{9}]_{0}^{1}

Compute the boundaries:

\frac{1}{9}

= \frac{1}{9}

\frac{e ^{x}}{ln ( x )}

\frac{d}{d x} (x t)

x \to - 2 - lim (\frac{1}{x - 2})

\frac{d}{d x} ((3 x^{4} - 4) (3 - 2 x^{3}))

\int x^{2} 1 + x d x

What is the integral from 0 to pi/4 of tan^8(θ)sec^2(θ) ?
The integral from 0 to pi/4 of tan^8(θ)sec^2(θ) is 1/9