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Question Number 142131 by Avijit007 last updated on 26/May/21

$$\int \frac{dx}{3+2sinx+cosx}dx$$

Commented by Avijit007 last updated on 26/May/21

$$help$$

Answered by 676597498 last updated on 26/May/21

$$y=\frac{1}{3+\frac{4t}{1+t^2}+\frac{1-t^2}{1+t^2}}=\frac{1+t^2}{3+3t^2+4t+1-t^2}$$$$\Rightarrow y=\frac{1+t^2}{2t^2+4t+4}=\frac{1}{2}\left(\frac{1+t^2}{t^2+2t+2}\right)$$$$t=tan\left(\frac{x}{2}\right)\Rightarrow dt=\frac{1}{2}(1+t^2)dx$$$$\Rightarrow dx=\frac{dt}{\frac{1}{2}(1+t^2)}$$$$I=\int \frac{dt}{{(t+1)}^2+1^2}$$$$(t+1)=tanu\Rightarrow dt={sec}^{2}udu$$$$\Rightarrow I=\int \frac{{sec}^{2}u}{{tan}^{2}u+1}=\int du=u+k$$$$I={tan}^{-1}(t+1)+k={tan}^{-1}(tan\left(\frac{x}{2}\right)+1)+k$$

Commented by Avijit007 last updated on 26/May/21

$$thankyou$$

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