What is the integral of tan^3(t)sec^2(t) ?

The integral of tan^3(t)sec^2(t) is (tan^4(t))/4+C

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integral of tan^3(t)sec^2(t)

\int tan^{3} (t) sec^{2} (t) d t

\frac{tan ^{4} ( t )}{4} + C

Solution steps

\int tan^{3} (t) sec^{2} (t) d t

Rewrite using trig identities

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

Apply u-substitution

: \int u^{3} d u

= \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

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

Substitute back

u = tan (t)

= \frac{tan ^{4} ( t )}{4}

Add a constant to the solution

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

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

\frac{s}{( s + 1 ) ^{2}}

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

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

\int x^{4} 4 + x^{5} d x

What is the integral of tan^3(t)sec^2(t) ?
The integral of tan^3(t)sec^2(t) is (tan^4(t))/4+C