What is the integral of sec^7(x)tan^3(x) ?

The integral of sec^7(x)tan^3(x) is (sec^9(x))/9-(sec^7(x))/7+C

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integral of sec^7(x)tan^3(x)

\int sec^{7} (x) tan^{3} (x) d x

\frac{sec ^{9} ( x )}{9} - \frac{sec ^{7} ( x )}{7} + C

Solution steps

\int sec^{7} (x) tan^{3} (x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (x)) tan (x) sec^{7} (x) d x

Apply u-substitution

: \int u^{6} (u^{2} - 1) d u

= \int u^{6} (u^{2} - 1) d u

Expand

u^{6} (u^{2} - 1) : u^{8} - u^{6}

= \int u^{8} - u^{6} 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^{8} d u - \int u^{6} d u

\int u^{8} d u = \frac{u ^{9}}{9}

\int u^{6} d u = \frac{u ^{7}}{7}

= \frac{u ^{9}}{9} - \frac{u ^{7}}{7}

Substitute back

u = sec (x)

= \frac{sec ^{9} ( x )}{9} - \frac{sec ^{7} ( x )}{7}

Add a constant to the solution

= \frac{sec ^{9} ( x )}{9} - \frac{sec ^{7} ( x )}{7} + C

x \frac{d y}{d x} = \frac{1}{y ^{2}} + y

- 3 sin^{4} (x)

\frac{\partial}{\partial x} (5 y^{4} cos (4 x))

\int sin^{\frac{1}{2}} (x) cos (x) d x

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

What is the integral of sec^7(x)tan^3(x) ?
The integral of sec^7(x)tan^3(x) is (sec^9(x))/9-(sec^7(x))/7+C