What is the integral of sec^4(x)tan^3(x) ?

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

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integral of sec^4(x)tan^3(x)

\int sec^{4} (x) tan^{3} (x) d x

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

Solution steps

\int sec^{4} (x) tan^{3} (x) d x

Rewrite using trig identities

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = tan (x)

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

Add a constant to the solution

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

\frac{1 - 2 s}{5 s ^{2} + 8 s + 10}

y^{^{'}} = t + t y^{2}

f (x) = x + x, at x = 4

\int_{- 1}^{4} (x^{2} + 2 x - 1) d x

\frac{d}{d x} (- y x)

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