What is the integral of 1/(sqrt(-t^2+8t-12)) ?

The integral of 1/(sqrt(-t^2+8t-12)) is arcsin(1/2 (t-4))+C

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integral of 1/(sqrt(-t^2+8t-12))

\int \frac{1}{- t ^{2} + 8 t - 12} d t

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

Solution steps

\int \frac{1}{- t ^{2} + 8 t - 12} d t

Complete the square

- t^{2} + 8 t - 12 : - (t - 4)^{2} + 4

= \int \frac{1}{- ( t - 4 ) ^{2} + 4} d t

Apply u-substitution

: \int \frac{1}{- u ^{2} + 4} d u

= \int \frac{1}{- u ^{2} + 4} 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}{2} (t - 4))

Simplify

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

Add a constant to the solution

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

(e^{x} + e^{- x}) (\frac{d y}{d x}) = y^{2}

x^{- 2}, 2

\int \frac{cos ^{3} ( x )}{sin ^{8} ( x )} d x

y = 7 x \cdot ln (5 x) - 7 x

\frac{d}{d x} (\frac{( x + 2 ) ^{2}}{1 - x})

What is the integral of 1/(sqrt(-t^2+8t-12)) ?
The integral of 1/(sqrt(-t^2+8t-12)) is arcsin(1/2 (t-4))+C