What is the integral of csc(t/4)cot(t/4) ?

The integral of csc(t/4)cot(t/4) is -4csc(t/4)+C

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integral of csc(t/4)cot(t/4)

\int csc (\frac{t}{4}) cot (\frac{t}{4}) d t

- 4 csc (\frac{t}{4}) + C

Solution steps

\int csc (\frac{t}{4}) cot (\frac{t}{4}) d t

Apply u-substitution

: \int csc (u) cot (u) \cdot 4 d u

= \int csc (u) cot (u) \cdot 4 d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int csc (u) cot (u) d u

Apply u-substitution

: \int - 1 d v

= 4 \cdot \int - 1 d v

Integral of a constant:

\int a d x = a x

= 4 (- 1) v

Substitute back

= 4 (- 1) csc (\frac{t}{4})

Simplify

= - 4 csc (\frac{t}{4})

Add a constant to the solution

= - 4 csc (\frac{t}{4}) + C

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

\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

What is the integral of csc(t/4)cot(t/4) ?
The integral of csc(t/4)cot(t/4) is -4csc(t/4)+C