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

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

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

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

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

Solution steps

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

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