What is the integral of 1/(sqrt(-t^2+4t+12)) ?

The integral of 1/(sqrt(-t^2+4t+12)) is arcsin(1/4 (t-2))+C

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integral of 1/(sqrt(-t^2+4t+12))

\int \frac{1}{- t ^{2} + 4 t + 12} d t

arcsin (\frac{1}{4} (t - 2)) + C

Solution steps

\int \frac{1}{- t ^{2} + 4 t + 12} d t

Complete the square

- t^{2} + 4 t + 12 : - (t - 2)^{2} + 16

= \int \frac{1}{- ( t - 2 ) ^{2} + 16} d t

Apply u-substitution

: \int \frac{1}{- u ^{2} + 16} d u

= \int \frac{1}{- u ^{2} + 16} d u

Apply Trigonometric Substitution

: \int 1 d v

= \int 1 d v

Integral of a constant:

\int a d x = a x

= 1 \cdot v

Substitute back

= 1 \cdot arcsin (\frac{1}{4} (t - 2))

Simplify

= arcsin (\frac{1}{4} (t - 2))

Add a constant to the solution

= arcsin (\frac{1}{4} (t - 2)) + C

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What is the integral of 1/(sqrt(-t^2+4t+12)) ?
The integral of 1/(sqrt(-t^2+4t+12)) is arcsin(1/4 (t-2))+C