Question

Evaluate: $\int\limits^{\pi/2}_0 {\frac{\cos x}{(4+\sin x)(3+\sin x)}} \, dx$

Answer 1

we have to evaluate the integration $\int\limits^{\pi/2}_0{\frac{\cos x}{(4+\sin x)(3+\sin x)}} \, dx$

let sinx = p........(1)

differentiating both sides,

cosx dx = dp.........(2)

lower limit, p = sin(0) = 0

upper limit , p = sin(π/2) = 1

putting equations (1) and (2) in above integration we get,

$\int\limits^1_0 {\frac{1}{(4+p)(3+p)}} \,dp$

= $\int\limits^1_0\left[\frac{1}{(3+p)}-\frac{1}{(4+p)}\right]\,dp$

= $[ln(3+p)]^1_0-[ln(4+p)]^1_0$

= (ln4 - ln3) - (ln5 - ln4)

= 2ln4 - ln3 - ln5

= ln(16) - (ln3 + ln5)

= ln(16) - ln(3 × 5)

= ln (16) - ln(15)

= ln(16/15)

Answer 2

Answer:

$\int\limits^{\pi/2}_0{\frac{\cos x}{(4+\sin x)(3+\sin x)}}dx= ln(16/15)$

Step-by-step explanation:

In the given question,

We have to integrate the term,

$\int\limits^{\pi/2}_0{\frac{\cos x}{(4+\sin x)(3+\sin x)}}dx$

Now,

Let us take the value of,

sin x = t

So,

On differentiating we get,

dt = cosx.dx

Also,

At, x = 0,

t = sin(0) = 0

And,

At, x = π/2

t = sin(π/2) = 1

On putting this in the given equation we get,

$=\int\limits^1_0 {\frac{1}{(4+t)(3+t)}}dt\\=\int\limits^1_0\left[\frac{1}{(3+t)}-\frac{1}{(4+t)}\right]dt\\= [ln(3+t)]^1_0-[ln(4+t)]^1_0$

On putting the limits we get,

$=(ln4-ln3)-(ln5-ln4)\\=2ln4-ln3-ln5 \\= ln(16) - (ln3 + ln5)\\= ln(16) - ln(15)\\= ln(16/15)$

Therefore, the final evaluated solution of the equation is given by,

$= ln(16/15)$