What is the integral of 6/(sqrt(3-2t-t^2)) ?

The integral of 6/(sqrt(3-2t-t^2)) is 6arcsin(1/2 (t+1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 6/(sqrt(3-2t-t^2))

\int \frac{6}{3 - 2 t - t ^{2}} d t

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

Solution steps

\int \frac{6}{3 - 2 t - t ^{2}} d t

Take the constant out:

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

= 6 \cdot \int \frac{1}{3 - 2 t - t ^{2}} d t

Complete the square

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

= 6 \cdot \int \frac{1}{- ( t + 1 ) ^{2} + 4} d t

Apply u-substitution

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

= 6 \cdot \int \frac{1}{- u ^{2} + 4} d u

Apply Trigonometric Substitution

: \int 1 d v

= 6 \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 6 \cdot 1 \cdot v

Substitute back

= 6 \cdot 1 \cdot arcsin (\frac{1}{2} (t + 1))

Simplify

= 6 arcsin (\frac{1}{2} (t + 1))

Add a constant to the solution

= 6 arcsin (\frac{1}{2} (t + 1)) + C

\int 8 x^{9} d x

\int s^{\frac{1}{5}} cos (s^{\frac{6}{5}} + 3) d s

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

\int (x^{3} + 8)^{5} 3 x^{2} d x

\int 2 e^{0.2 x} d x

What is the integral of 6/(sqrt(3-2t-t^2)) ?
The integral of 6/(sqrt(3-2t-t^2)) is 6arcsin(1/2 (t+1))+C