Integral Calculus Question 285
Question: $ \int_{{}}^{{}}{\frac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}}\ dx= $
Options:
A) $ \tan x+\cot x+c $
B) $ \tan x-\cot x+c $
C) $ cosec,x-\cot x+c $
D) $ \sec x-cosec,x+c $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \int_{{}}^{{}}{\frac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{2}}x{{\cos }^{2}}x},dx}=\int_{{}}^{{}}{( \frac{\sin x}{{{\cos }^{2}}x}+\frac{\cos x}{{{\sin }^{2}}x} ),dx} $ $ =\frac{{{(x-5)}^{-2+1}}}{-2+1}+c=\frac{{{(x-5)}^{-1}}}{-1}+c $ $ =\sec x-\cos ecx+c $
BETA
