What is the integral of 1/(sqrt(27-6θ-θ^2)) ?

The integral of 1/(sqrt(27-6θ-θ^2)) is arcsin(1/6 (θ+3))+C

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integral of 1/(sqrt(27-6θ-θ^2))

\int \frac{1}{27 - 6 θ - θ ^{2}} d θ

arcsin (\frac{1}{6} (θ + 3)) + C

Solution steps

\int \frac{1}{27 - 6 θ - θ ^{2}} d θ

Complete the square

27 - 6 θ - θ^{2} : - (θ + 3)^{2} + 36

= \int \frac{1}{- ( θ + 3 ) ^{2} + 36} d θ

Apply u-substitution

: \int \frac{1}{- u ^{2} + 36} d u

= \int \frac{1}{- u ^{2} + 36} 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}{6} (θ + 3))

Simplify

= arcsin (\frac{1}{6} (θ + 3))

Add a constant to the solution

= arcsin (\frac{1}{6} (θ + 3)) + C

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What is the integral of 1/(sqrt(27-6θ-θ^2)) ?
The integral of 1/(sqrt(27-6θ-θ^2)) is arcsin(1/6 (θ+3))+C