What is the integral of (10)/(sqrt(9-16x^2)) ?

The integral of (10)/(sqrt(9-16x^2)) is 5/2 arcsin(4/3 x)+C

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integral of (10)/(sqrt(9-16x^2))

\int \frac{10}{9 - 16 x ^{2}} d x

\frac{5}{2} arcsin (\frac{4}{3} x) + C

Solution steps

\int \frac{10}{9 - 16 x ^{2}} d x

Take the constant out:

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

= 10 \cdot \int \frac{1}{9 - 16 x ^{2}} d x

Apply Trigonometric Substitution

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

= 10 \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

= 10 \cdot \frac{1}{4} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 10 \cdot \frac{1}{4} \cdot 1 \cdot u

Substitute back

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

= 10 \cdot \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{3} x)

Simplify

10 \cdot \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{3} x) : \frac{5}{2} arcsin (\frac{4}{3} x)

= \frac{5}{2} arcsin (\frac{4}{3} x)

Add a constant to the solution

= \frac{5}{2} arcsin (\frac{4}{3} x) + C

\int \frac{1}{x ^{2} + 4 x - 12} d x

y^{^{''}} + 6 y^{^{'}} + 8 y = 0, y (0) = 9

cos (e^{- 4 θ^{5}})

\frac{d}{d x} (\frac{9 x ^{2} - 6}{x ^{2}})

\frac{\partial}{\partial x} (\frac{4}{1 + x ^{2}})

What is the integral of (10)/(sqrt(9-16x^2)) ?
The integral of (10)/(sqrt(9-16x^2)) is 5/2 arcsin(4/3 x)+C