Question
If $\int \frac{x^{2}+20}{(x \sin x+5 \cos x)^{2}} d x=-\frac{x}{A \cos x}+B$, then(A) $A=x \sin x+5 \cos x$(B) $B=\cot x$(C) $A=-(x \sin x+5 \cos x)$(D) $B=\tan x$
Step 1
This gives us: \[\int \frac{x^{3}+20x}{(x \sin x+5 \cos x)^{2}} dx\] Show more…
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