What is the integral of 7/(sqrt(16-4t^2)) ?

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

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integral of 7/(sqrt(16-4t^2))

\int \frac{7}{16 - 4 t ^{2}} d t

\frac{7}{2} arcsin (\frac{t}{2}) + C

Solution steps

\int \frac{7}{16 - 4 t ^{2}} d t

Take the constant out:

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

= 7 \cdot \int \frac{1}{16 - 4 t ^{2}} d t

Apply Trigonometric Substitution

: \int \frac{1}{2} d u

= 7 \cdot \int \frac{1}{2} d u

Take the constant out:

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

= 7 \cdot \frac{1}{2} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 7 \cdot \frac{1}{2} \cdot 1 \cdot u

Substitute back

u = arcsin (\frac{2}{4} t)

= 7 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{4} t)

Simplify

7 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{4} t) : \frac{7}{2} arcsin (\frac{t}{2})

= \frac{7}{2} arcsin (\frac{t}{2})

Add a constant to the solution

= \frac{7}{2} arcsin (\frac{t}{2}) + C

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