Mathematics

$$\int {\dfrac{{(3\sin x - 2)\cos x}}{{5 - {{\cos }^2}x - 4\sin x}}dx} $$


SOLUTION
$$I=\int \dfrac{(3\sin x-2)\cos x}{5-cos^2x-4\sin x}dx$$

$$\Rightarrow$$  $$\dfrac{(3\sin x-2)}{\sin^2x-4\sin x+4}dx$$

$$\Rightarrow$$  $$\dfrac{(3\sin x-2)\cos x}{(\sin x-2)^2}dx$$

Let $$\sin x=t$$
So, $$\cos x\, dx=dt$$
$$\therefore$$  $$I=\int\dfrac{(3t-2)}{(t-2)^2}dt$$

$$\Rightarrow$$  $$2\int\dfrac{tdt}{(t-2)^2}-2\int\dfrac{dt}{(t-2)^2}$$

$$\Rightarrow$$  $$3\left[t\left(\dfrac{1}{-3(t-2)^3}\right)-\int\dfrac{dt}{-3(t-2)^3}\right]-2\left(\dfrac{1}{-3(t-2)^3}\right)+c$$

$$\Rightarrow$$  $$\dfrac{-t}{(t-2)^3}+\dfrac{1}{3}\left[\dfrac{1}{-4(t-2)^4}\right]+\dfrac{2}{3(t-2)^3}+c$$

$$\Rightarrow$$  $$\dfrac{1}{(t-2)^3}\left(-t+\dfrac{2}{3}\right)-\dfrac{1}{12(t-2)^4}+c$$

$$\Rightarrow$$  $$-\dfrac{1}{(t-2)^2}-\dfrac{4}{3(t-2)^3}-\dfrac{1}{12(t-2)^4}+c$$

$$\Rightarrow$$  $$-\dfrac{1}{(\sin x-2)^2}-\dfrac{4}{3(\sin x-2)^3}-\dfrac{1}{12\sin x-2)^4}+c$$


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Subjective Medium Published on 17th 09, 2020
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