What is the integral of 5/(4sin(x)+3cos(x)) ?

The integral of 5/(4sin(x)+3cos(x)) is ln| 1/5+3/5 tan(x/2)|-ln|-9/5+3/5 tan(x/2)|+C

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integral of 5/(4sin(x)+3cos(x))

\int \frac{5}{4 sin ( x ) + 3 cos ( x )} d x

ln \frac{1}{5} + \frac{3}{5} tan (\frac{x}{2}) - ln - \frac{9}{5} + \frac{3}{5} tan (\frac{x}{2}) + C

Solution steps

\int \frac{5}{4 sin ( x ) + 3 cos ( 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}{4 sin ( x ) + 3 cos ( x )} d x

Apply u-substitution

: \int \frac{2}{8 u + 3 ( 1 - u ^{2} )} d u

= 5 \cdot \int \frac{2}{8 u + 3 ( 1 - u ^{2} )} d u

Take the constant out:

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

= 5 \cdot 2 \cdot \int \frac{1}{8 u + 3 ( 1 - u ^{2} )} d u

Complete the square

8 u + 3 (1 - u^{2}) : - 3 (u - \frac{4}{3})^{2} + \frac{25}{3}

= 5 \cdot 2 \cdot \int \frac{1}{- 3 ( u - \frac{4}{3} ) ^{2} + \frac{25}{3}} d u

Apply u-substitution

: \int \frac{3}{- 9 v ^{2} + 25} d v

= 5 \cdot 2 \cdot \int \frac{3}{- 9 v ^{2} + 25} d v

Take the constant out:

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

= 5 \cdot 2 \cdot 3 \cdot \int \frac{1}{- 9 v ^{2} + 25} d v

Apply Integral Substitution

= 5 \cdot 2 \cdot 3 \cdot \int \frac{1}{15 ( - w ^{2} + 1 )} d w

Take the constant out:

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

= 5 \cdot 2 \cdot 3 \cdot \frac{1}{15} \cdot \int \frac{1}{- w ^{2} + 1} d w

Use the common integral:

\int \frac{1}{- w ^{2} + 1} d w = \frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}

= 5 \cdot 2 \cdot 3 \cdot \frac{1}{15} (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2})

Substitute back

= 5 \cdot 2 \cdot 3 \cdot \frac{1}{15} (\frac{ln \frac{3}{5} ( tan ( \frac{x}{2} ) - \frac{4}{3} ) + 1}{2} - \frac{ln \frac{3}{5} ( tan ( \frac{x}{2} ) - \frac{4}{3} ) - 1}{2})

Simplify

5 \cdot 2 \cdot 3 \cdot \frac{1}{15} (\frac{ln \frac{3}{5} ( tan ( \frac{x}{2} ) - \frac{4}{3} ) + 1}{2} - \frac{ln \frac{3}{5} ( tan ( \frac{x}{2} ) - \frac{4}{3} ) - 1}{2}) : ln \frac{1}{5} + \frac{3}{5} tan (\frac{x}{2}) - ln - \frac{9}{5} + \frac{3}{5} tan (\frac{x}{2})

= ln \frac{1}{5} + \frac{3}{5} tan (\frac{x}{2}) - ln - \frac{9}{5} + \frac{3}{5} tan (\frac{x}{2})

Add a constant to the solution

= ln \frac{1}{5} + \frac{3}{5} tan (\frac{x}{2}) - ln - \frac{9}{5} + \frac{3}{5} tan (\frac{x}{2}) + C

x^{2} y^{^{''}} + 2 x y^{^{'}} - 6 y = 0, y (1) = 0, y^{^{'}} (1) = - 4

\frac{d}{d x} (sin (ln (4) x^{2}))

y = x 400 - x^{2}

\frac{d}{d x} (x^{5} sin (2 x))

What is the integral of 5/(4sin(x)+3cos(x)) ?
The integral of 5/(4sin(x)+3cos(x)) is ln| 1/5+3/5 tan(x/2)|-ln|-9/5+3/5 tan(x/2)|+C