_ ^ x x a ^2 x + b ^2 x dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{\sin x\cos x}}{{a{{\cos }^2}x + b{{\sin }^2}x}}dx = } $

✓
$\frac{1}{{2(b - a)}}\log (a{\cos ^2}x + b{\sin ^2}x) + c$
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

✓

Answer

Correct option: A.

$\frac{1}{{2(b - a)}}\log (a{\cos ^2}x + b{\sin ^2}x) + c$

a
(a) 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$.

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