What is the integral of sec^4(5x)tan^3(5x) ?

The integral of sec^4(5x)tan^3(5x) is (sec^6(5x))/(30)-(sec^4(5x))/(20)+C

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integral of sec^4(5x)tan^3(5x)

\int sec^{4} (5 x) tan^{3} (5 x) d x

\frac{sec ^{6} ( 5 x )}{30} - \frac{sec ^{4} ( 5 x )}{20} + C

Solution steps

\int sec^{4} (5 x) tan^{3} (5 x) d x

Rewrite using trig identities

= \int (- 1 + sec^{2} (5 x)) tan (5 x) sec^{4} (5 x) d x

Apply u-substitution

: \int \frac{u ^{3} ( u ^{2} - 1 )}{5} d u

= \int \frac{u ^{3} ( u ^{2} - 1 )}{5} d u

Expand

\frac{u ^{3} ( u ^{2} - 1 )}{5} : \frac{u ^{5}}{5} - \frac{u ^{3}}{5}

= \int \frac{u ^{5}}{5} - \frac{u ^{3}}{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 \frac{u ^{5}}{5} d u - \int \frac{u ^{3}}{5} d u

\int \frac{u ^{5}}{5} d u = \frac{u ^{6}}{30}

\int \frac{u ^{3}}{5} d u = \frac{u ^{4}}{20}

= \frac{u ^{6}}{30} - \frac{u ^{4}}{20}

Substitute back

u = sec (5 x)

= \frac{sec ^{6} ( 5 x )}{30} - \frac{sec ^{4} ( 5 x )}{20}

Add a constant to the solution

= \frac{sec ^{6} ( 5 x )}{30} - \frac{sec ^{4} ( 5 x )}{20} + C

f^{^{'}} (x) = \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + x}}}

\frac{x ^{3}}{1 - x ^{2}}

\int (\frac{3 x ^{2} + 2}{x ^{2}}) d x

f (x) = x sin (x), 0, 2 π

What is the integral of sec^4(5x)tan^3(5x) ?
The integral of sec^4(5x)tan^3(5x) is (sec^6(5x))/(30)-(sec^4(5x))/(20)+C