What is the integral of 5/(16+9cos^2(x)) ?

The integral of 5/(16+9cos^2(x)) is 1/4 arctan(4/5 tan(x))+C

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integral of 5/(16+9cos^2(x))

\int \frac{5}{16 + 9 cos ^{2} ( x )} d x

\frac{1}{4} arctan (\frac{4}{5} tan (x)) + C

Solution steps

\int \frac{5}{16 + 9 cos ^{2} ( x )} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{16 + 9 cos ^{2} ( x )} d x

Apply Trigonometric Substitution

: \int \frac{1}{16 u ^{2} + 25} d u

= 5 \cdot \int \frac{1}{16 u ^{2} + 25} d u

Apply Integral Substitution

= 5 \cdot \int \frac{1}{20 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 5 \cdot \frac{1}{20} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= 5 \cdot \frac{1}{20} arctan (v)

Substitute back

= 5 \cdot \frac{1}{20} arctan (\frac{4}{5} tan (x))

Simplify

5 \cdot \frac{1}{20} arctan (\frac{4}{5} tan (x)) : \frac{1}{4} arctan (\frac{4}{5} tan (x))

= \frac{1}{4} arctan (\frac{4}{5} tan (x))

Add a constant to the solution

= \frac{1}{4} arctan (\frac{4}{5} tan (x)) + C

\int \frac{tan ^{3} ( x )}{cos ^{4} ( x )} d x

\frac{\partial}{\partial x} (ln ((\frac{y}{x})^{2}))

\int \frac{x ^{2} - 8 x - 17}{( x - 1 ) ^{2} ( x ^{2} + 1 )} d x

\frac{\partial}{\partial x} (41 - x^{2} - y^{2})

\frac{\partial}{\partial x} (sin (2 y))

What is the integral of 5/(16+9cos^2(x)) ?
The integral of 5/(16+9cos^2(x)) is 1/4 arctan(4/5 tan(x))+C