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INTEGRALS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS5 Marks
Question
$\int\limits^{\frac{\pi}{2}}_0\frac{\text{dx}}{(\text{a}^2\cos^2\text{x}+\text{b}^2\sin^2\text{x})^2}$
Hint: Divide Numerator and Denominator by $\cos^4x$
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Answer

Let $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\text{dx}}{(\text{a}^2\cos^2\text{x}+\text{b}^2\sin^2\text{x})^2}$
Divide numerator and denominator by $\cos^4x$, we get
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sec^4\text{x dx}}{(\text{a}^2+\text{b}^2\tan^2\text{x})^2}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{(1+\tan^2\text{x})\sec^4\text{x dx}}{(\text{a}^2+\text{b}^2\tan^2\text{x})^2}$
Put $\tan\text{x}=\text{t}$
$\Rightarrow\ \sec^2\text{x dx}=\text{dt}$
As $\text{x}\rightarrow0,$ then $\text{t}\rightarrow0$
and $\text{x}\rightarrow\frac{\pi}{2}$ then$\text{t}\rightarrow\infty$
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{(1+\text{t}^2)}{(\text{a}^2+\text{b}^2\text{t}^2)^2}$
Now, $\frac{1+\text{t}^2}{(\text{a}^2+\text{b}^2\text{t}^2)^2}$ $[\text{let t}^2=\text{u}]$
$\frac{1+\text{u}}{(\text{a}^2+\text{b}^2\text{u})^2}=\frac{\text{A}}{(\text{a}^2+\text{b}^2\text{u})}+\frac{\text{B}}{(\text{a}^2+\text{b}^2\text{u})^2}$
$\Rightarrow\ 1+\text{u}=\text{A}(\text{a}^2+\text{b}^2\text{u})\text{B}$
On comparing the coefficient of x and constant term on both sides, we get
$\text{a}^2\text{A}+\text{B}=1\ \ \dots(\text{i})$
and $\text{b}^2\text{A}=1\ \ \dots(\text{ii})$
$\therefore\ \text{A}=\frac{1}{\text{b}^2}$
Now, $\frac{\text{a}^2}{\text{b}^2}+\text{B}=1$
$\text{B}=1-\frac{\text{a}^2}{\text{b}^2}=\frac{\text{b}^2-\text{a}^2}{\text{b}^2}$
$\text{I}\int\limits^\infty_0\frac{(1+\text{t}^2)}{(\text{a}^2+\text{b}^2\text{t}^2)^2}$
$=\frac{1}{\text{b}^2}\int\limits^\infty_0\frac{\text{dt}}{\text{a}^2+\text{b}^2\text{t}^2}+\frac{\text{b}^2-\text{a}^2}{\text{b}^2}\int\limits^\infty_0\frac{\text{dt}}{(\text{a}^2+\text{b}^2\text{t}^2)^2}$
$=\frac{1}{\text{b}^2}\int\limits^\infty_0\frac{\text{dt}}{\text{b}^2\Big(\frac{\text{a}^2}{\text{b}^2}+\text{t}^2\Big)}+\frac{\text{b}^2-\text{a}^2}{\text{b}^2}\int\limits^\infty_0\frac{\text{dt}}{(\text{a}^2+\text{b}^2\text{t}^2)^2}$
$=\frac{1}{\text{ab}^2}\bigg[\tan^{-1}\Big(\frac{\text{tb}}{\text{a}}\Big)\bigg]^\infty_0+\frac{\text{b}^2-\text{a}^2}{\text{b}^2}\Big(\frac{\pi}{4}\cdot\frac{1}{\text{a}^3\text{b}}\Big)$
$=\frac{1}{\text{ab}^3}\big[\tan^{-1}\infty-\tan^{-1}0\big]+\frac{\pi}{4}\cdot\frac{\text{b}^2-\text{a}^2}{(\text{a}^3\text{b}^3)}$
$=\frac{\pi}{2\text{ab}^3}+\frac{\pi}{4}\cdot\frac{\text{b}^2-\text{a}^2}{(\text{a}^3\text{b}^3)}$
$=\pi\Big(\frac{2\text{a}^2+\text{b}^2-\text{a}^2}{4\text{a}^3\text{b}^3}\Big)=\frac{\pi}{4}\Big(\frac{\text{a}^2+\text{b}^2}{\text{a}^3\text{b}^3}\Big)$

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