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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int\frac{1}{\cos\text{x}+\text{cosec x}}\text{dx}$
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Answer

Let $\text{I}=\int\frac{1}{\cos\text{x}+\text{cosec x}}\text{dx}$
$=\int\frac{\sin\text{x}}{1+\cos\text{x}\sin\text{x}}\text{dx}$
$=\int\frac{2\sin\text{x}}{2+2\cos\text{x}\sin\text{x}}\text{dx}$
$=\int\frac{\sin\text{x}+\cos\text{x}+\sin\text{x}-\cos\text{x}}{2+2\cos\text{x}\sin\text{x}}\text{dx}$
$=\int\frac{\sin\text{x}+\cos\text{x}}{2+2\cos\text{x}\sin\text{x}}\text{dx}+\int\frac{\sin\text{x}-\cos\text{x}}{2+2\cos\text{x}\sin\text{x}}\text{dx}$
$=\int\frac{\sin\text{x}+\cos\text{x}}{3-\sin^2\text{x}-\cos^2\text{x}+2\cos\text{x}\sin\text{x}}\text{dx}+\int\frac{\sin\text{x}-\cos\text{x}}{1+\sin^2\text{x}+\cos^2\text{x}+2\cos\text{x}\sin\text{x}}\text{dx}$
$=\int\frac{\sin\text{x}+\cos\text{x}}{3-(\sin\text{x}-\cos\text{x})^2}\text{dx}+\int\frac{\sin\text{x}-\cos\text{x}}{1+(\sin\text{x}+\cos\text{x})^2}\text{dx}$
$=\text{I}_1+\text{I}_2 \dots(1)$
Where, $\text{I}_1=\int\frac{\sin\text{x}+\cos\text{x}}{3-(\sin\text{x}-\cos\text{x}^2)}\text{dx}$ and $\text{I}_2=\int\frac{\sin\text{x}-\cos\text{x}}{1+(\sin\text{x}+\cos\text{x})^2}\text{dx}$
Now,
$\text{I}_1=\int\frac{\sin\text{x}+\cos\text{x}}{3-(\sin\text{x}-\cos\text{x})^2}\text{dx}$
Let $(\sin\text{x}-\cos\text{x})=\text{t}$
On differentiating both sides, we get
$(\cos\text{x}+\sin\text{x})\text{dx = dt}$
$\therefore\text{I}_1=\int\frac{1}{3-(\text{t})^2}\text{dt}$
$=\frac{1}{2\sqrt{3}}\log\bigg|\frac{\sqrt{3}+\text{t}}{\sqrt{3}-\text{t}}\bigg|+\text{C}_1$
$=\frac{1}{2\sqrt{3}}\log\bigg|\frac{\sqrt{3}+\sin\text{x}-\cos\text{x}}{\sqrt{3}-(\sin\text{x}-\cos\text{x})}\bigg|+\text{C}_1 \dots(2)$
Now,
$\text{I}_2=\int\frac{\sin\text{x}-\cos\text{x}}{1+(\sin\text{x}+\cos\text{x})^2}\text{dx}$
Let $(\sin\text{x}+\cos\text{x})=\text{t}$
On differentiating both sides, we get
$(\cos\text{x}-\sin\text{x})\text{dx = dt}$
$\therefore\text{I}_2=-\int\frac{1}{1+(\text{t})^2}\text{dt}$
$=-\tan^{-1}\text{t}+\text{c}_2$
$=-\tan^{-1}(\sin\text{x}+\cos\text{x})+\text{c}_2 \dots(3)$
On substituting (2) and (3) in (1), we get
$\text{I}=\frac{1}{2\sqrt{3}}\log\bigg|\frac{\sqrt{3}+\sin\text{x}-\cos\text{x}}{\sqrt{3}-\sin\text{x}+\cos\text{x}}\bigg|-\tan^{-1}(\sin\text{x}+\cos\text{x})+\text{C}$
Hence, $\int\frac{1}{\cos\text{x}+\text{cosec x}}\text{dx}=\frac{1}{2\sqrt{3}}\log\bigg|\frac{\sqrt{3}+\sin\text{x}-\cos\text{x}}{\sqrt{3}-\sin\text{x}+\cos\text{x}}\bigg|-\tan^{-1}(\sin\text{x}+\cos\text{x})+\text{C}$

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