What is the integral of 1/(7cos(x)-4sin(x)+8) ?

The integral of 1/(7cos(x)-4sin(x)+8) is -ln|tan(x/2)-3|+ln|tan(x/2)-5|+C

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integral of 1/(7cos(x)-4sin(x)+8)

\int \frac{1}{7 cos ( x ) - 4 sin ( x ) + 8} d x

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

Solution steps

\int \frac{1}{7 cos ( x ) - 4 sin ( x ) + 8} d x

Apply u-substitution

: \int \frac{2}{u ^{2} - 8 u + 15} d u

= \int \frac{2}{u ^{2} - 8 u + 15} d u

Take the constant out:

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

= 2 \cdot \int \frac{1}{u ^{2} - 8 u + 15} d u

Take the partial fraction of

\frac{1}{u ^{2} - 8 u + 15} : - \frac{1}{2 ( u - 3 )} + \frac{1}{2 ( u - 5 )}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{2 ( u - 3 )} d u = \frac{1}{2} ln ∣ u - 3 ∣

\int \frac{1}{2 ( u - 5 )} d u = \frac{1}{2} ln ∣ u - 5 ∣

= 2 (- \frac{1}{2} ln ∣ u - 3 ∣ + \frac{1}{2} ln ∣ u - 5 ∣)

Substitute back

u = tan (\frac{x}{2})

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 1/(7cos(x)-4sin(x)+8) ?
The integral of 1/(7cos(x)-4sin(x)+8) is -ln|tan(x/2)-3|+ln|tan(x/2)-5|+C