What is the integral of sec^4(x)tan^4(x) ?

The integral of sec^4(x)tan^4(x) is (tan^5(x))/5+(tan^7(x))/7+C

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integral of sec^4(x)tan^4(x)

\int sec^{4} (x) tan^{4} (x) d x

\frac{tan ^{5} ( x )}{5} + \frac{tan ^{7} ( x )}{7} + C

Solution steps

\int sec^{4} (x) tan^{4} (x) d x

Rewrite using trig identities

= \int (1 + tan^{2} (x)) sec^{2} (x) tan^{4} (x) d x

Apply u-substitution

: \int u^{4} (1 + u^{2}) d u

= \int u^{4} (1 + u^{2}) d u

Expand

u^{4} (1 + u^{2}) : u^{4} + u^{6}

= \int u^{4} + 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^{4} d u + \int u^{6} d u

\int u^{4} d u = \frac{u ^{5}}{5}

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

= \frac{u ^{5}}{5} + \frac{u ^{7}}{7}

Substitute back

u = tan (x)

= \frac{tan ^{5} ( x )}{5} + \frac{tan ^{7} ( x )}{7}

Add a constant to the solution

= \frac{tan ^{5} ( x )}{5} + \frac{tan ^{7} ( x )}{7} + C

f (x) = (x^{3} + 7 x^{2} + 5)^{5}

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

\frac{d}{d x} (cos (e^{x^{2}}))

\int \frac{sin ( x )}{2 x} d x

, x

What is the integral of sec^4(x)tan^4(x) ?
The integral of sec^4(x)tan^4(x) is (tan^5(x))/5+(tan^7(x))/7+C