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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 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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