What is the integral of 7/(sqrt(1-49x^2)) ?

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

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integral of 7/(sqrt(1-49x^2))

\int \frac{7}{1 - 49 x ^{2}} d x

arcsin (7 x) + C

Solution steps

\int \frac{7}{1 - 49 x ^{2}} d x

Take the constant out:

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

= 7 \cdot \int \frac{1}{1 - 49 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{7} d u

= 7 \cdot \int \frac{1}{7} d u

Take the constant out:

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

= 7 \cdot \frac{1}{7} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 7 \cdot \frac{1}{7} \cdot 1 \cdot u

Substitute back

u = arcsin (7 x)

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

Simplify

7 \cdot \frac{1}{7} \cdot 1 \cdot arcsin (7 x) : arcsin (7 x)

= arcsin (7 x)

Add a constant to the solution

= arcsin (7 x) + C

\int (2 x - 1)^{7} d x

\int \frac{1}{2 v} d v

\int \frac{x}{2 x + 1} d x

\int (2 x^{2} + 4)^{8} \cdot x d x

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