What is the integral of (sin(x))/(cos^2(x)+cos(x)-6) ?

The integral of (sin(x))/(cos^2(x)+cos(x)-6) is -2(1/10 ln|3tan^2(x/2)+1|-1/10 ln|tan^2(x/2)+2|)+C

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integral of (sin(x))/(cos^2(x)+cos(x)-6)

\int \frac{sin ( x )}{cos ^{2} ( x ) + cos ( x ) - 6} d x

- 2 (\frac{1}{10} ln 3 tan^{2} (\frac{x}{2}) + 1 - \frac{1}{10} ln tan^{2} (\frac{x}{2}) + 2) + C

Solution steps

\int \frac{sin ( x )}{cos ^{2} ( x ) + cos ( x ) - 6} d x

Rewrite using trig identities

= \int \frac{2 sin ( x )}{cos ( 2 x ) + 2 cos ( x ) - 11} d x

Take the constant out:

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

= 2 \cdot \int \frac{sin ( x )}{cos ( 2 x ) + 2 cos ( x ) - 11} d x

Rewrite using trig identities

= 2 \cdot \int \frac{sin ( x )}{2 cos ^{2} ( x ) + 2 cos ( x ) - 12} d x

Apply u-substitution

: \int - \frac{u}{3 u ^{4} + 7 u ^{2} + 2} d u

= 2 \cdot \int - \frac{u}{3 u ^{4} + 7 u ^{2} + 2} d u

Take the constant out:

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

= 2 (- \int \frac{u}{3 u ^{4} + 7 u ^{2} + 2} d u)

Take the partial fraction of

\frac{u}{3 u ^{4} + 7 u ^{2} + 2} : \frac{3 u}{5 ( 3 u ^{2} + 1 )} - \frac{u}{5 ( u ^{2} + 2 )}

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

\int \frac{3 u}{5 ( 3 u ^{2} + 1 )} d u = \frac{1}{10} ln 3 u^{2} + 1

\int \frac{u}{5 ( u ^{2} + 2 )} d u = \frac{1}{10} ln u^{2} + 2

= 2 (- (\frac{1}{10} ln 3 u^{2} + 1 - \frac{1}{10} ln u^{2} + 2))

Substitute back

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

= 2 (- (\frac{1}{10} ln 3 tan^{2} (\frac{x}{2}) + 1 - \frac{1}{10} ln tan^{2} (\frac{x}{2}) + 2))

Simplify

= - 2 (\frac{1}{10} ln 3 tan^{2} (\frac{x}{2}) + 1 - \frac{1}{10} ln tan^{2} (\frac{x}{2}) + 2)

Add a constant to the solution

= - 2 (\frac{1}{10} ln 3 tan^{2} (\frac{x}{2}) + 1 - \frac{1}{10} ln tan^{2} (\frac{x}{2}) + 2) + C

f (x) = x^{2} - 3 x + 5, at x = 1

n = 0 \sum \infty n^{3}

f (x) = - 7 x^{3} - 8 x^{2} - 5 x, at x = - 3

\int \frac{sin ( 2 x )}{1 + cos ( 2 x )} d x

\int \frac{1}{( 7 x - 1 ) 49 x ^{2} - 14 x} d x

What is the integral of (sin(x))/(cos^2(x)+cos(x)-6) ?
The integral of (sin(x))/(cos^2(x)+cos(x)-6) is -2(1/10 ln|3tan^2(x/2)+1|-1/10 ln|tan^2(x/2)+2|)+C