Question

integrate of (1/sinx+root 3cosx)dx​

Answer 1

Answer:

Please see the attachment

Answer 2

Answer:

           I= $\frac{1}{2}$ ㏒ tan $(\frac{x}{2} +\frac{\pi }{6})$ + C

Given: Given that (1/sinx+root 3cosx)dx​

To find: We have to find the integral of (1/sinx+root 3cosx)dx​

Explanation:

• For solving the problem first we have to multiply both numerator and denominator  of the equation (1/sinx+root 3cosx)dx​ by 1/2.
• Therefore integral of the equation (1/sinx+root 3cosx)dx,

                             I= $\frac{1}{2}$ $\int\ {\frac{1}{\frac{1}{2}sin x +\frac{\sqrt{3} }{2}cos x }} \, dx$

                        ⇒  I= $\frac{1}{2}$ $\int\ {\frac{1}{cos \frac{\pi }{3} sin x + sin \frac{\pi }{3} cos x } } \, dx$

• Now the equation inside the integral part is in the form of $\frac{1}{sin (A+B)}$
• Hence,   I= $\frac{1}{2}$ $\int {\frac{1}{sin (x+\frac{\pi }{3}) } } \, dx$
• We know that the reciprocal of sine function is cosec function
• Then I= $\frac{1}{2}$ $\int\ {cosec (x+\frac{\pi }{3} )} \, dx$

• We know $\int\ {cosec x} \, dx$= ㏒ tan(x/2) + C

• Therefore I= $\frac{1}{2}$ ㏒ tan $(\frac{x}{2} +\frac{\pi }{6})$ + C
• So that the final answer $\frac{1}{2}$ ㏒ tan $(\frac{x}{2} +\frac{\pi }{6})$ + C is the integral of (1/sinx+root 3cosx)dx​, where C is the universal constant of integration.

To know more about the concept please go through the links:

https://brainly.in/question/41180374

https://brainly.in/question/7692308

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