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

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

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integral of tan^8(θ)sec^2(θ)

\int tan^{8} (θ) sec^{2} (θ) d θ

\frac{tan ^{9} ( θ )}{9} + C

Solution steps

\int tan^{8} (θ) sec^{2} (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

Apply the Power Rule

= \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

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