What is the integral of sec^2(θ)tan^5(θ) ?

The integral of sec^2(θ)tan^5(θ) is (tan^6(θ))/6+C

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integral of sec^2(θ)tan^5(θ)

\int sec^{2} (θ) tan^{5} (θ) d θ

\frac{tan ^{6} ( θ )}{6} + C

Solution steps

\int sec^{2} (θ) tan^{5} (θ) d θ

Rewrite using trig identities

= \int (1 + tan^{2} (θ)) tan^{5} (θ) d θ

Apply u-substitution

: \int u^{5} d u

= \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= \frac{u ^{6}}{6}

Substitute back

u = tan (θ)

= \frac{tan ^{6} ( θ )}{6}

Add a constant to the solution

= \frac{tan ^{6} ( θ )}{6} + C

\int \frac{x}{x ^{2} + 4 x + 8} d x

\int arccot (9 x) d x

y^{^{''}} - 2 y^{^{'}} - 15 y = 3 x cos (2 x)

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

x \to \infty lim (- 2 e^{\frac{2}{x}})

What is the integral of sec^2(θ)tan^5(θ) ?
The integral of sec^2(θ)tan^5(θ) is (tan^6(θ))/6+C