Question

π/2∫ cosx/(1+sinx)(2+sinx)(3+sinx) dx ,Evaluate it.0

Answer 1

Answer:

ecostheta + esintneta

Answer 2

Answer:

㏑$\frac{4\sqrt{2} }{3\sqrt{3} }$

Step-by-step explanation:

We have to find the value of,

$\int\limits^\frac{\pi }{2} _0 {\frac{Cosx}{(1+Sinx)(2+Sinx)(3+Sinx)} } \, dx$.

Let us first calculate the indefinite integral. Then we will put the limits.

∫$\frac{Cosx}{(1+Sinx)(2+Sinx)(3+Sinx)}dx$

Assume that Sinx = z, ⇒Cosx.dx=dz. So, the above integration can be written as,

∫$\frac{dz}{(1+z)(2+z)(3+z)}$

=∫$\frac{(2+z)-(1+z)}{(1+z)(2+z)(3+z)}dz$

=∫$\frac{dz}{(1+z)(3+z)}$-∫$\frac{dz}{(2+z)(3+z)}$

= $\frac{1}{2} \int\ {\frac{(3+z)-(1+z)}{(3+z)(1+z)} } \, dz-\int\ {\frac{(3+z)-(2+z)}{(3+z)(2+z)} } \, dz$

= $\frac{1}{2}\int\ {\frac{1}{1+z} } \, dz-\frac{1}{2}\int\ {\frac{1}{3+z} } \, dz-\int\ {\frac{1}{2+z} } \, dz+\int\ {\frac{1}{3+z} } \, dz$

= $\frac{1}{2}\int\ {\frac{1}{1+z} } \, dz-\int\ {\frac{1}{2+z} } \, dz+\frac{1}{2}\int\ {\frac{1}{3+z} } \, dz$

=$\frac{1}{2}$㏑[1+z]-㏑[2+z]+$\frac{1}{2}$㏑[3+z] +c {Where C is the integration constant.}

= ㏑$\frac{\sqrt{(1+z)(3+z)}}{2+z}$ +c

Now putting z=Sinx, we will get,

=㏑$\frac{\sqrt{(1+Sinx)(3+Sinx)}}{2+Sinx}$ +c

So, putting limits x=0 to x=π/2, we will get

=㏑$\frac{\sqrt{(1+1)(3+1)} }{2+1}$-㏑$\frac{\sqrt{(1+0)(3+0)} }{2+0}$

=㏑$\frac{2\sqrt{2} }{3}$-㏑$\frac{\sqrt{3} }{2}$

=㏑$\frac{4\sqrt{2} }{3\sqrt{3} }$

(Answer)