What is the integral of cot(t)sqrt(sin^2(t)-25) ?

The integral of cot(t)sqrt(sin^2(t)-25) is -5arcsec(1/5 sin(t))+sqrt(sin^2(t)-25)+C

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integral of cot(t)sqrt(sin^2(t)-25)

\int cot (t) sin^{2} (t) - 25 d t

- 5 arcsec (\frac{1}{5} sin (t)) + sin^{2} (t) - 25 + C

Solution steps

\int cot (t) sin^{2} (t) - 25 d t

Apply u-substitution

: \int \frac{u ^{2} - 25}{u} d u

= \int \frac{u ^{2} - 25}{u} d u

Apply Trigonometric Substitution

: \int 5 tan^{2} (v) d v

= \int 5 tan^{2} (v) d v

Take the constant out:

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

= 5 \cdot \int tan^{2} (v) d v

Rewrite using trig identities

= 5 \cdot \int - 1 + sec^{2} (v) d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 5 (- \int 1 d v + \int sec^{2} (v) d v)

\int 1 d v = v

\int sec^{2} (v) d v = tan (v)

= 5 (- v + tan (v))

Substitute back

= 5 (- arcsec (\frac{1}{5} sin (t)) + tan (arcsec (\frac{1}{5} sin (t))))

Simplify

5 (- arcsec (\frac{1}{5} sin (t)) + tan (arcsec (\frac{1}{5} sin (t)))) : - 5 arcsec (\frac{1}{5} sin (t)) + sin^{2} (t) - 25

= - 5 arcsec (\frac{1}{5} sin (t)) + sin^{2} (t) - 25

Add a constant to the solution

= - 5 arcsec (\frac{1}{5} sin (t)) + sin^{2} (t) - 25 + C

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

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\frac{d}{d x} (ln (2 x - x))

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\int (\frac{5}{x} + \frac{15}{x ^{6}}) d x

What is the integral of cot(t)sqrt(sin^2(t)-25) ?
The integral of cot(t)sqrt(sin^2(t)-25) is -5arcsec(1/5 sin(t))+sqrt(sin^2(t)-25)+C