What is the integral of (7/(sqrt(9-16x^2))) ?

The integral of (7/(sqrt(9-16x^2))) is 7/4 arcsin(4/3 x)+C

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integral of (7/(sqrt(9-16x^2)))

\int (\frac{7}{9 - 16 x ^{2}}) d x

\frac{7}{4} arcsin (\frac{4}{3} x) + C

Solution steps

\int \frac{7}{9 - 16 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}{9 - 16 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{4} d u

= 7 \cdot \int \frac{1}{4} d u

Take the constant out:

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

= 7 \cdot \frac{1}{4} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 7 \cdot \frac{1}{4} \cdot 1 \cdot u

Substitute back

u = arcsin (\frac{4}{3} x)

= 7 \cdot \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{3} x)

Simplify

7 \cdot \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{3} x) : \frac{7}{4} arcsin (\frac{4}{3} x)

= \frac{7}{4} arcsin (\frac{4}{3} x)

Add a constant to the solution

= \frac{7}{4} arcsin (\frac{4}{3} x) + C

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What is the integral of (7/(sqrt(9-16x^2))) ?
The integral of (7/(sqrt(9-16x^2))) is 7/4 arcsin(4/3 x)+C