Let I=∫0π/2cosxacosx+bsinxdxJ=∫0π/2sinxacosx+bsinxdx, a>0, b>0 Which of the following is/are true?

aI+bJ=π2 bI−aJ=∫0π/2(bcosx−asinxacosx+bsinx)dx =ln(ba) ⇒ a2I+abJ=aπ2 b2I−abJ= b ln(ba)I=1a2+b2(aπ2+bln(ba)) & J=1a2+b2(bπ2−aln(ba))

Let I=∫0π/2cosxacosx+bsinxdxJ=∫0π/2sinxacosx+bsinxdx, a>0, b>0 Which of the following is/are true?

aI+bJ=π2 bI−aJ=∫0π/2(bcosx−asinxacosx+bsinx)dx =ln(ba) ⇒ a2I+abJ=aπ2 b2I−abJ= b ln(ba)I=1a2+b2(aπ2+bln(ba)) & J=1a2+b2(bπ2−aln(ba))

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Let $I = \int_{0}^{π / 2} \frac{\cos x}{a \cos x + b \sin x} d x$
$J = \int_{0}^{π / 2} \frac{\sin x}{a \cos x + b \sin x} d x, a > 0, b > 0$
Which of the following is/are true?

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$I = \frac{1}{a^{2} + b^{2}} (\frac{a π}{2} + b \ln (\frac{b}{a}))$

$J = \frac{1}{a^{2} + b^{2}} (\frac{b π}{2} + a \ln (\frac{b}{a}))$

$I = \frac{1}{a^{2} + b^{2}} (\frac{b π}{2} - a \ln (\frac{b}{a}))$

$J = \frac{1}{a^{2} + b^{2}} (\frac{a π}{2} - b \ln (\frac{b}{a}))$

answer is A.

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Detailed Solution

$a I + b J = \frac{π}{2} b I - a J = \int_{0}^{π / 2} (\frac{b \cos x - a \sin x}{a \cos x + b \sin x}) d x = \ln (\frac{b}{a}) \Rightarrow a^{2} I + a b J = \frac{a π}{2} \frac{b^{2} I - a b J = b \ln (\frac{b}{a})}{I = \frac{1}{a^{2} + b^{2}} (\frac{a π}{2} + b \ln (\frac{b}{a}))} & J = \frac{1}{a^{2} + b^{2}} (\frac{b π}{2} - a \ln (\frac{b}{a}))$