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Question Number 63117 by mugan deni last updated on 29/Jun/19

$$\int \frac{\cos x}{2+3\sin x+\sin {}^{2}x}dx$$

Commented by mathmax by abdo last updated on 30/Jun/19

$$letI=\int \frac{cosx}{{sin}^{2}x+3sinx+2}dxchangementsinx=tgivecosxdx=dt\Rightarrow$$$$I=\int \frac{dt}{t^2+3t+2}letsolve{t}^2+3t+2=0\to \mathrm{\Delta }=9-8=1\Rightarrow$$$$t_1=\frac{-3+1}{2}=-1and{t}_2=\frac{-3-1}{2}=-2\Rightarrow t^2+3t+2=(t+1)(t+2)\Rightarrow$$$$I=\int \frac{dt}{(t+1)(t+2)}=\int (\frac{1}{t+1}-\frac{1}{t+2})dt=ln\mid t+1\mid -ln\mid t+2\mid +c$$$$=ln\mid 1+sinx\mid -ln\mid 2+sinx\mid +c.$$

Answered by Hope last updated on 29/Jun/19

$$a=sinxda=cosxdx$$$$\int \frac{da}{a^2+3a+2}$$$$\int \frac{da}{(a+1)(a+2)}$$$$\int \frac{(a+2)-(a+1)}{(a+1)(a+2)}da$$$$\int \frac{da}{a+1}-\int \frac{da}{a+2}$$$$ln(a+1)-ln(a+2)+c$$$$ln\left(\frac{a+1}{a+2}\right)+c$$$$ln\left(\frac{1+sinx}{2+sinx}\right)+c$$$$$$$$$$

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