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

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

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integral of sec^2(x)tan^5(x)

\int sec^{2} (x) tan^{5} (x) d x

\frac{tan ^{6} ( x )}{6} + C

Solution steps

\int sec^{2} (x) tan^{5} (x) d x

Rewrite using trig identities

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

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 (x)

= \frac{tan ^{6} ( x )}{6}

Add a constant to the solution

= \frac{tan ^{6} ( x )}{6} + C

\frac{d y}{d x} + 2 x y = - x

t y^{^{'}} + 8 y = t^{2} - t + 1

\frac{d}{d x} (\frac{1}{( 25 - x ^{2} )})

\int e^{7} d x

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 t e^{2 t}, y (0) = 1, y^{^{'}} (0) = 0

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