Question

Integration π/3 to 0 cosx/3+4sinx.dx

Answer 1

Please find the attached image of solution.

Answer 2

Answer:

$\dfrac{1}{4}log(\dfrac{3}{3+2\sqrt{3}})$

Step-by-step explanation:

We have to evaluate the integration of

$\dfrac{cosx}{3+4sinx}$ from $\dfrac{\pi}{3}$ to 0.

Let's assume that

$y\ =\ \int^{0}_{\frac{\pi}{3}}{\dfrac{cosx}{3+4sinx}dx}$           (1)

Let's assume that

     t = sinx

=>dt = cosx dx

When x = 0

=> t = 0

when $x\ =\ \dfrac{\pi}{3}$

$=>\ t\ =\ sin\dfrac{\pi}{3}$

         $=\ \dfrac{\sqrt{3}}{2}$

By putting these values in equation (1), we will get

      $y\ =\ \int^{0}_{\frac{\sqrt{3}}{2}}\dfrac{dt}{3+4t}}$

          $=\ \int^0_{\frac{\sqrt{3}}{2}}\dfrac{dt}{4(t+\dfrac{3}{4})}$

          $=\ \dfrac{1}{4}[log{t+\dfrac{3}{4}}]^0_{\frac{\sqrt{3}}{2}}$

          $=\ \dfrac{1}{4}[log\dfrac{3}{4}-log(\dfrac{\sqrt{3}}{2}+\dfrac{3}{4})]$

          $=\ \dfrac{1}{4}[log\dfrac{3}{4}-log\dfrac{2\sqrt{3}+3}{4}]$

          $=\ \dfrac{1}{4}[log\dfrac{3\times 4}{4\times (2\sqrt{3}+3)}]$

          $=\ \dfrac{1}{4}[log\dfrac{3}{2\sqrt{3}+3}]$

Hence, the integration of the given term is $\dfrac{1}{4}[log\dfrac{3}{2\sqrt{3}+3}]$