What is the integral of tan^3(x)sec^6(x) ?

The integral of tan^3(x)sec^6(x) is (tan^4(x))/4+(tan^6(x))/3+(tan^8(x))/8+C

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integral of tan^3(x)sec^6(x)

\int tan^{3} (x) sec^{6} (x) d x

\frac{tan ^{4} ( x )}{4} + \frac{tan ^{6} ( x )}{3} + \frac{tan ^{8} ( x )}{8} + C

Solution steps

\int tan^{3} (x) sec^{6} (x) d x

Rewrite using trig identities

= \int sec^{2} (x) (1 + tan^{2} (x))^{2} tan^{3} (x) d x

Apply u-substitution

: \int u^{3} (1 + u^{2})^{2} d u

= \int u^{3} (1 + u^{2})^{2} d u

Expand

u^{3} (1 + u^{2})^{2} : u^{3} + 2 u^{5} + u^{7}

= \int u^{3} + 2 u^{5} + u^{7} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u^{3} d u + \int 2 u^{5} d u + \int u^{7} d u

\int u^{3} d u = \frac{u ^{4}}{4}

\int 2 u^{5} d u = \frac{u ^{6}}{3}

\int u^{7} d u = \frac{u ^{8}}{8}

= \frac{u ^{4}}{4} + \frac{u ^{6}}{3} + \frac{u ^{8}}{8}

Substitute back

u = tan (x)

= \frac{tan ^{4} ( x )}{4} + \frac{tan ^{6} ( x )}{3} + \frac{tan ^{8} ( x )}{8}

Add a constant to the solution

= \frac{tan ^{4} ( x )}{4} + \frac{tan ^{6} ( x )}{3} + \frac{tan ^{8} ( x )}{8} + C

\int \frac{1}{9} x^{- \frac{2}{9}} d x

\frac{\partial}{\partial x} (y^{2} \cdot x)

\int_{1}^{9} \frac{1}{2 x ^{2}} d x

x^{2} y^{^{''}} + 2 x y^{^{'}} - 12 y = 0, y (1) = - 1, y^{^{'}} (1) = 10

\frac{d x}{d y} = - \frac{4 y ^{2} + 6 x y}{3 y ^{2} + 2 x}

What is the integral of tan^3(x)sec^6(x) ?
The integral of tan^3(x)sec^6(x) is (tan^4(x))/4+(tan^6(x))/3+(tan^8(x))/8+C