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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Evalute the following integrals:
$\int\frac{\sin2\text{x}}{\text{a}\cos^2\text{x}+\text{b}\sin^2\text{x}}\text{dx}$
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Answer

Let $\text{I}=\int\frac{\sin2\text{x}}{\text{a}\cos^2\text{x}+\text{b}\sin^2\text{x}}\text{dx}$

$=\int\frac{\sin2\text{x}}{\text{a}(1-\sin^2\text{x})+\text{b}\sin^2\text{x}}\text{dx}$

$=\int\frac{\sin2\text{x}}{(\text{b}-\text{a})\sin^2\text{x}+\text{a}}\text{dx}$

Putting $\sin^2\text{x}=\text{t}$

$\Rightarrow2\sin\text{x}\cos\text{x}=\frac{\text{dt}}{\text{dx}}$

$\Rightarrow\sin2\text{x}=\frac{\text{dt}}{\text{dx}}$

$\Rightarrow\sin2\text{x }\text{dx}=\text{dt}$

$\therefore\text{I}=\int\frac{1}{(\text{b}-\text{a})\text{t}+\text{a}}\text{dt}$

$=\frac{1}{(\text{b}-\text{a})}\text{ln}|(\text{b}-\text{a})\text{t}+\text{a}|+\text{C}$

$\Big[\because\int\frac{1}{\text{ax}+\text{b}}\text{dx}=\frac{1}{\text{a}}\text{In}|\text{ax}+\text{b}|+\text{C}\Big]$

$=\frac{1}{(\text{b}-\text{a})}\text{ln}|(\text{b}-\text{a})\sin^2\text{x}+\text{a}|+\text{C}\ \big[\because\text{t}=\sin^2\text{x}\big]$

$=\frac{1}{(\text{b}-\text{a})}\text{ln}\big|\text{b}\sin^2\text{x}+\text{a}(1-\sin^2\text{x})\big|+\text{C}$

$=\frac{1}{(\text{b}-\text{a})}\text{ln}\big|\text{b}\sin^2\text{x}+\text{a}\cos^2\text{x}\big|+\text{C}$

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