What is the integral of 6/(sqrt(49-x^2)) ?

The integral of 6/(sqrt(49-x^2)) is 6arcsin(1/7 x)+C

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integral of 6/(sqrt(49-x^2))

\int \frac{6}{49 - x ^{2}} d x

6 arcsin (\frac{1}{7} x) + C

Solution steps

\int \frac{6}{49 - x ^{2}} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{49 - x ^{2}} d x

Apply Trigonometric Substitution

: \int 1 d u

= 6 \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 6 \cdot 1 \cdot u

Substitute back

u = arcsin (\frac{1}{7} x)

= 6 \cdot 1 \cdot arcsin (\frac{1}{7} x)

Simplify

= 6 arcsin (\frac{1}{7} x)

Add a constant to the solution

= 6 arcsin (\frac{1}{7} x) + C

\int \frac{arcsin ( \frac{1}{x} )}{x ^{5}} d x

\int \frac{- 5}{2 x} d x

\int \frac{x ^{2}}{( 49 + x ^{2} ) ^{2}} d x

\int 1 - sec^{2} (x) d x

What is the integral of 6/(sqrt(49-x^2)) ?
The integral of 6/(sqrt(49-x^2)) is 6arcsin(1/7 x)+C