If I= displaystyle ∫ (sin x/3 sin â¡ x + cos â¡ x + 2)dx and J= displaystyle ∫ (cos x/3 sin â¡ x + cos â¡ x + 2)dx , then 3J-I is equal to (where C is the constant of integration)

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Tardigrade Admin · Accepted Answer

3J-I= displaystyle ∫ (3 c o s x - s i n x/3 s i n x + c o s x + 2)dx Let 3sinx+cosx+2=t ⇒ (3 cos x - sin â¡ x)dx=dt So, 3J-I= displaystyle ∫ (d t/t)=ln |t|+C 3J-I=ln |3 sin â¡ x + cos â¡ x + 2|+c

Q. If $I = \int \frac{s in x}{3 s in x + cos x + 2} d x$ and $J = \int \frac{cos x}{3 s in x + cos x + 2} d x$ , then $3 J - I$ is equal to (where $C$ is the constant of integration)

$x + l n ∣ 3 s in x + cos x + 2 ∣ + c$

$x - l n ∣ 3 s in x + cos x + 2 ∣ + c$

$l n ∣ 3 s in x + cos x + 2 ∣ + c$

$2 x + l n ∣ 3 s in x + cos x + 2 ∣ + c$

$3 J - I = \int \frac{3 cos x - s in x}{3 s in x + cos x + 2} d x$
Let $3 s in x + cos x + 2 = t$
$\Rightarrow (3 cos x - s in x) d x = d t$
So, $3 J - I = \int \frac{d t}{t} = l n ∣ t ∣ + C$
$3 J - I = l n ∣ 3 s in x + cos x + 2 ∣ + c$

Q. If I=∫3sin⁡x+cos⁡x+2sinx​dx and J=∫3sin⁡x+cos⁡x+2cosx​dx , then 3J−I is equal to (where C is the constant of integration)

x+ln∣3sin⁡x+cos⁡x+2∣+c

x−ln∣3sin⁡x+cos⁡x+2∣+c

ln∣3sin⁡x+cos⁡x+2∣+c

2x+ln∣3sin⁡x+cos⁡x+2∣+c

3J−I=∫3sinx+cosx+23cosx−sinx​dx Let 3sinx+cosx+2=t ⇒(3cosx−sin⁡x)dx=dt So, 3J−I=∫tdt​=ln∣t∣+C 3J−I=ln∣3sin⁡x+cos⁡x+2∣+c

Q. If $I = \int \frac{s in x}{3 s in x + cos x + 2} d x$ and $J = \int \frac{cos x}{3 s in x + cos x + 2} d x$ , then $3 J - I$ is equal to (where $C$ is the constant of integration)

$x + l n ∣ 3 s in x + cos x + 2 ∣ + c$

$x - l n ∣ 3 s in x + cos x + 2 ∣ + c$

$l n ∣ 3 s in x + cos x + 2 ∣ + c$

$2 x + l n ∣ 3 s in x + cos x + 2 ∣ + c$

$3 J - I = \int \frac{3 cos x - s in x}{3 s in x + cos x + 2} d x$
Let $3 s in x + cos x + 2 = t$
$\Rightarrow (3 cos x - s in x) d x = d t$
So, $3 J - I = \int \frac{d t}{t} = l n ∣ t ∣ + C$
$3 J - I = l n ∣ 3 s in x + cos x + 2 ∣ + c$