Mathematics

Evaluate :
$$\displaystyle\int {\dfrac{9cosx-sinx}{4sinx+5cosx}dx}$$


ANSWER

$$x+\ln(4\sin x+5\cos x)+c$$


SOLUTION
$$\to \displaystyle \int \dfrac{9\cos x - \sin x}{4\sin x + 5 \cos x} dx$$
It is in the form $$\dfrac{f(x)}{g(x)}$$, so we will write $$f(x)$$ as $$A(g(x)) + B(g'(x))$$.
$$\to f(x) = 9\cos x - \sin x$$
$$\to g(x) = 4\sin x + 5 \cos x$$
    $$g'(x) = 4 \cos x - 5 \sin x$$
So, $$f(x) = g(x) + g'(x)$$
$$\to \displaystyle \int 1+ \dfrac{g'(x)}{g(x)} dx \Rightarrow x + ln (g(x))+C$$
Answer $$= x + ln (4\sin x + 5\cos x) + C$$ 
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Single Correct Medium Published on 17th 09, 2020
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