Integral Calculus Question 356
Question: $ \int_{{}}^{{}}{\frac{\sin x\cos x}{a{{\cos }^{2}}x+b{{\sin }^{2}}x}dx=} $
[AI CBSE 1988, 89]
Options:
A) $ \frac{1}{2(b-a)}\log (a{{\cos }^{2}}x+b{{\sin }^{2}}x)+c $
B) $ \frac{1}{b-a}\log (a{{\cos }^{2}}x+b{{\sin }^{2}}x)+c $
C) $ \frac{1}{2}\log (a{{\cos }^{2}}x+b{{\sin }^{2}}x)+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Put $ a{{\cos }^{2}}x+b{{\sin }^{2}}x=t $
$ \Rightarrow 2(b-a)\sin x\cos x=dt, $ then $ \int_{{}}^{{}}{\frac{\sin x\cos x,dx}{a{{\cos }^{2}}x+b{{\sin }^{2}}x}=\frac{1}{2(b-a)}\int_{{}}^{{}}{\frac{1}{t},dt}} $ $ =\frac{1}{2(b-a)}\log (a{{\cos }^{2}}x+b{{\sin }^{2}}x)+c $ .
BETA
