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INTEGRALS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS5 Marks
Question
Prove that $\int\limits_{0}^{\text{a}}\text{f(x)}\text{dx}=\int\limits_{0}^{\text{a}}\text{f}(\text{a}-\text{x})\text{dx},$ hence evaluate $\int\limits_{0}^\pi\frac{\text{x}\sin\text{x}}{1+\cos^2\text{x}}\text{dx}.$
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Answer

To prove: $\int\limits_{0}^{\text{a}}\text{f(x)}\text{dx}=\int\limits_{0}^{\text{a}}\text{f}(\text{a}-\text{x})\text{dx}$
Proof: Let
$\text{Let} \ \text{a}-\text{x}=\text{t}$
$\Rightarrow\text{dx}=-\text{dt}$
When $\text{x}=0,\text{t}=\text{a}$
When $\text{x}=\text{a},\text{t}=0$
Putting the value of x in LHS
$\int\limits_\text{a}^{0}\text{f}(\text{a}-\text{t})(-\text{dt})$
$=-\int\limits_\text{a}^{0}\text{f}(\text{a}-\text{t})(\text{dt})$
$=\int\limits_\text{a}^{0}\text{f}(\text{a}-\text{t})(\text{dt})$
 $\Big(\therefore\int\limits_\text{a}^\text{b}\text{f(t)}\text{d}\text{t}=\int\limits_\text{a}^\text{b}\text{f(x)dx}\Big)$
$\text{LHS}=\text{RHS}$
Using this we can solve the given question as follows:
$=\int\limits_0^\pi\frac{\text{x}\sin\text{x}}{1+\cos^2\text{x}}\text{dx}=\text{I}\dots(1)$
On Putting $\text{x}=\pi-\text{x}$
$\int\limits_0^\pi\frac{(\pi-\text{x})\sin(\pi-\text{x})}{1+\cos^2(\pi-\text{x})}=\text{I}\dots(2)$
$\text{I}+\text{I}=\int\limits_{0}^{\pi}\frac{\text{x}\sin\text{x}\text{ dx}}{1+\cos^{2}\text{x}}+\int\limits^{\pi}_0\frac{(\pi-\text{x})\sin(\pi-\text{x})}{1+\cos^2(\pi-\text{x})}$
$\Rightarrow2\text{I}=\int\limits_0^\pi\frac{\text{x}\sin\text{x}}{1+\cos^2\text{x}}\text{dx}+0$
$\Rightarrow2\text{I}=\int\limits_0^\pi\frac{\text{x}\sin\text{x}}{1+\cos^2\text{x}}\text{dx}$
$\text{Let}\cos\text{x}=0$
$-\sin\text{x}\text{ dx}=\text{dv}$
$\Rightarrow2\text{I}=-\text{x}\int\limits^{\pi}_{\text{0}}\frac{\text{dv}}{1+\text{v}^2}$
$\Rightarrow2\text{I}=-\pi\int\limits^{-1}_1\frac{1}{1+\text{v}^2}\text{dv}$ $[\text{Given}\text{ x}=\pi(\text{limit})]$
$\Rightarrow2\text{I}=-\pi[\tan^{-1}\text{v}]_1^{-1}$
$\Rightarrow2\text{I}=-\pi\Big[-\frac{\pi}{4}-\frac{\pi}{4}\Big]=\frac{\pi^{2}}{2}$
$\text{I}=\frac{\pi^{2}}{4}$

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