What is the integral of 1/(sqrt(-t^2+2t+3)) ?

The integral of 1/(sqrt(-t^2+2t+3)) is arcsin(1/2 (t-1))+C

Popular >

integral of 1/(sqrt(-t^2+2t+3))

\int \frac{1}{- t ^{2} + 2 t + 3} d t

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

Solution steps

\int \frac{1}{- t ^{2} + 2 t + 3} d t

Complete the square

- t^{2} + 2 t + 3 : - (t - 1)^{2} + 4

= \int \frac{1}{- ( t - 1 ) ^{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 - 1))

Simplify

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

Add a constant to the solution

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

4 x^{2} + y = 4, x^{8} - y = 1

\int_{1}^{5} (x - 1)^{2} d x

\int \frac{12}{x ^{2}} d x

\frac{d}{d x} (x - \frac{1}{3} x)

\frac{\partial}{\partial x} (x e^{6 y} sin (5 z))

What is the integral of 1/(sqrt(-t^2+2t+3)) ?
The integral of 1/(sqrt(-t^2+2t+3)) is arcsin(1/2 (t-1))+C