What is the integral of sin(t)sec^5(t) ?

The integral of sin(t)sec^5(t) is (sec^4(t))/4+C

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integral of sin(t)sec^5(t)

\int sin (t) sec^{5} (t) d t

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

Solution steps

\int sin (t) sec^{5} (t) d t

Rewrite using trig identities

= \int tan (t) sec^{4} (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 = sec (t)

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

Add a constant to the solution

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

x \to 0 lim ((x) x)

\int arccos (\frac{1 + x}{2}) d x

\int \frac{1}{1 + u ^{2}} d u

y = x^{2}, y = x + 6

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

What is the integral of sin(t)sec^5(t) ?
The integral of sin(t)sec^5(t) is (sec^4(t))/4+C