If $\int \frac{s i n 2 x}{a^{2} c o s^{2} x + b^{2} {sin}^{2} x} d x =$ $k . log | a^{2} {cos}^{2} x + b^{2} {sin}^{2} x | + c$ ,
then $k =$

\frac{1}{b^{2} - a^{2}}

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\frac{1}{(b^{2} - a^{2})^{2}}

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\frac{1}{a^{2} - b^{2}}

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\frac{1}{a^{2} + b^{2}}

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Solution

The correct option is A $\frac{1}{b^{2} - a^{2}}$
Let $a^{2} {cos}^{2} x + b^{2} {sin}^{2} x = t$
$[a^{2} (2 cos x) (- sin x) + b^{2} (2 sin x) (cos x)] d x$
$= d t$
$[(b^{2} - a^{2}) sin 2 x]^{d x} = d t$
sin $2 x d x = \frac{d t}{b^{2} - a^{2}}$
$1 / b^{2} - a^{2} \int \frac{d t}{t} = \frac{1}{b^{2} - a^{2}} l o t t + c$
$t = a^{2} {cos}^{2} x + b^{2} {sin}^{2} x$
$\Rightarrow \frac{1}{b^{2} - a^{2}} log (a^{2} {cos}^{2} x + b^{2} {sin}^{2} x) + c$
So, $k = \frac{1}{b^{2} - a^{2}}$

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Similar questions

a + b + c = 0

\frac{1}{b^{2} + c^{2} - a^{2}} + \frac{1}{c^{2} + a^{2} - b^{2}} + \frac{1}{a^{2} + b^{2} - c^{2}}

a, b, c

a^{2} = b^{2} + c^{2} - 2 b c \sqrt{1 - a^{2}}, b^{2} = c^{2} + a^{2} - 2 c a \sqrt{1 - b^{2}}, c^{2} = a^{2} + b^{2} - 2 a b \sqrt{1 - c^{2}}

a = c \sqrt{1 - b^{2}} + b \sqrt{1 - c^{2}}

x = \frac{\sqrt{a^{2} + b^{2}} + \sqrt{a^{2} - b^{2}}}{\sqrt{a^{2} + b^{2}} - \sqrt{a^{2} - b^{2}}}

a^{2} - b^{2} = \frac{2 a^{2}}{x^{2} + 1}

x = \frac{\sqrt{a^{2} + b^{2}} + \sqrt{a^{2} - b^{2}}}{\sqrt{a^{2} + b^{2}} - \sqrt{a^{2} - b^{2}}}

b^{2} = \frac{2 a^{2} x}{x^{2} + 1}

$\frac{c o s 2 A}{a^{2}} - \frac{c o s 2 B}{b^{2}} = \frac{1}{a^{2}} - \frac{1}{b^{2}}$

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