What is the integral of tan^4(x)sec^2(x) ?

The integral of tan^4(x)sec^2(x) is (tan^5(x))/5+C

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integral of tan^4(x)sec^2(x)

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int u^{4} d u

= \int u^{4} d u

Apply the Power Rule

: \frac{u ^{5}}{5}

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

Substitute back

u = tan (x)

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

Add a constant to the solution

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

f (x) = x^{2} - 6 x, at x = - 2

\frac{d y}{d x} = \frac{( x ^{2} )}{y ( 1 + x ^{3} )}

\frac{d}{d x} (\frac{3}{( 1 - x ^{2} ) ^{4}})

What is the integral of tan^4(x)sec^2(x) ?
The integral of tan^4(x)sec^2(x) is (tan^5(x))/5+C